Evaluate the indefinite integral. ∫ − ( 6 ) x d x \int_{}^{}-\left(6\right)^{x}dx

− ( 6 ) x ln ⁡ 6 + C -\frac{\left(6\right)^{x}}{\ln6}+C

− x 6 x − 1 + C -x6^{x-1}+C

− 1 x + 1 6 x + 1 + C -\frac{1}{x+1}6^{x+1}+C

− ( 6 ) x + 1 ln ⁡ x + C -\frac{\left(6\right)^{x+1}}{\ln x}+C

Evaluate the indefinite integral. ∫ 3 x 4 − ( 5 ) x d x \int_{}^{}3x^4-\left(5\right)^{x}dx

3 5 x 5 − 5 x ln ⁡ 5 + C \frac35x^5-\frac{5^{x}}{\ln5}+C

3 5 x 5 − 5 x + 1 ln ⁡ 5 + C \frac35x^5-\frac{5^{x+1}}{\ln5}+C

12 x 5 − 5 x ln ⁡ 5 + C 12x^5-\frac{5^{x}}{\ln5}+C

12 x 5 − 5 x + 1 ln ⁡ 5 + C 12x^5-\frac{5^{x+1}}{\ln5}+C

Evaluate the indefinite integral. ∫ 2 ( 1 3 ) x d x \int_{}^{}2\left(\frac13\right)^{x}dx

− 2 3 x ln ⁡ 3 + C -\frac{2}{3^{x}\ln3}+C

− 2 x x + 1 4 + C -\frac{2x^{x+1}}{4}+C

− 2 x ln ⁡ 3 + C -\frac{2x}{\ln3}+C

− 2 x x + 1 3 x ln ⁡ 4 + C -\frac{2x^{x+1}}{3^{x}\ln4}+C

Evaluate the indefinite integral. ∫ ( x − e x ) d x \int_{}^{}\left(\sqrt{x}-e^{x}\right)dx

2 3 x 3 2 − e x + C \frac23x^{\frac32}-e^{x}+C

2 3 x 2 3 − e x + C \frac23x^{\frac23}-e^{x}+C

2 3 x 3 2 − x e x + C \frac23x^{\frac32}-xe^{x}+C

2 3 x 2 3 − 1 x e x + C \frac23x^{\frac23}-\frac{1}{x}e^{x}+C

Evaluate the indefinite integral. ∫ ( 4 e x + 1 x 3 ) d x \int_{}^{}\left(4e^{x}+\frac{1}{x^3}\right)dx

4 e x − 1 2 x 2 + C 4e^{x}-\frac{1}{2x^2}+C

4 e x − 1 4 x 4 + C 4e^{x}-\frac{1}{4x^4}+C

4 x e x − 1 2 x 2 + C 4xe^{x}-\frac{1}{2x^2}+C

4 x e x − 1 4 x 4 + C 4xe^{x}-\frac{1}{4x^4}+C

Evaluate the indefinite integral. ∫ e 2 x x d x \int_{}^{}\frac{e^{2\sqrt{x}}}{\sqrt{x}}dx

5 e 2 x 3 x 3 2 + C \frac{5e^{2\sqrt{x}}}{3x^{\frac32}}+C

3 e x 7 2 7 x 3 2 + C \frac{3e^{x^{\frac72}}}{7x^{\frac32}}+C

7 e x 7 2 3 x 3 2 + C \frac{7e^{x^{\frac72}}}{3x^{\frac32}}+C

Evaluate the indefinite integral. ∫ e sec ⁡ θ sec ⁡ θ tan ⁡ θ d θ \int_{}^{}e^{\sec\theta}\sec\theta\tan\theta d\theta

e sec ⁡ θ + C e^{\sec\theta}+C

− e sec ⁡ θ c o s 2 ⁡ θ + C -\frac{e^{\sec\theta}}{cos^2⁡\theta}+C

− e sec ⁡ θ + C -e^{\sec\theta}+C

e sec ⁡ θ c o s 2 ⁡ θ + C \frac{e^{\sec\theta}}{cos^2⁡\theta}+C

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

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To find the integral of an exponential function $b^{x}$ , divide by the natural log of the base: $\frac{b^{x}}{\ln (b)}$ plus a constant. For $e^{x}$ , the integral is $e^{x}$ itself. Use rules like the constant multiple rule and sum rule to simplify integrals. This approach is crucial for understanding integrals in calculus.

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concept

Integrals of General Exponential Functions

Video duration:

Video transcript

At this point in the course, we've learned a ton of different rules for finding the integrals of various functions. And we've learned a lot of those rules based on the idea that taking an integral is really just the reverse process of taking a derivative. And now that we're faced with finding the integral of our general exponential function, $ b^x $, we're going to use that same idea here. So let's go ahead and jump right into things. Now recall that when finding the derivative of our general exponential function, again, that's $ b^x $, we multiplied $ b^x $ by the natural log of $ b $.

So when finding the derivative of something like $ 7^x $, this gives us $ 7^x \times \ln(7) $. So now when faced with finding the integral of functions like this $ 7^x $, we just need to reverse this process. So when finding the integral of any general exponential function, we are now going to be dividing $ b^x $ by the natural log of $ b $ instead of multiplying. So coming down to our rule box here, when we take the integral of $ b^x \, dx $, this is going to give us $ \frac{b^x}{\ln(b)} $. Then, of course, since this is an indefinite integral, we need to add on that constant of integration $ c $.

Now looking at this rule, we have certain restrictions on our base $ b $ that it has to be greater than zero and not equal to one, but these are just restrictions based on the definition of a general exponential function and aren't typically something you're going to have to worry about. So let's go ahead and apply this rule to our integral here when we integrate $ 7^x \, dx $. Now here, I can see that I have this base of seven. So with $ b $ equal to seven, applying my rule here, this is going to give me $ \frac{7^x}{\ln(7)} $. Then, of course, plus that constant of integration, $ c $.

And that's all there is to it. Having reversed our rule for finding the derivative to now find the integral. Now, we can use all of the rules for integrals that we've already learned along with this new rule, so let's go ahead and work through one additional example. Here we're asked to evaluate the indefinite integral of $ 3 \times \left(\frac{1}{2}\right)^x + 8^x \, dx $. Now depending on how comfortable you are with integrals at this point, you may be able to do this all in one go.

But I'm going to go ahead and break this down step by step just so you can actually see what I'm doing along the way. Now because I have two different functions being added together here, I'm going to go ahead and split this into two integrals being added together using my sum rule. So this is going to be equal to the integral of $ 3 \times \left(\frac{1}{2}\right)^x \, dx $, and then plus the integral of $ 8^x \, dx $. Now I can do one more thing here to simplify this. I can use my constant multiple rule to pull that three out front of this integral, making this three times the integral of $ \left(\frac{1}{2}\right)^x \, dx $.

So now let's go ahead and find what this integral is. Now looking at this first term here, taking that constant, I'm going to multiply three times the integral of $ \left(\frac{1}{2}\right)^x \, dx $. So here I can see that this integral has a $ b $ value equal to one half. So when applying this rule here, I'm just plugging in that one half everywhere that $ b $ is. So this is going to be equal to $ \frac{\left(\frac{1}{2}\right)^x}{\ln\left(\frac{1}{2}\right)} $.

Then adding that together with my other integral here, here I can see that I have a base of eight. So plugging eight in where $ b $ is for my rule here, this is going to give me $ \frac{8^x}{\ln(8)} $. Then because we're working with indefinite integrals here, I need to go ahead and add on that constant of integration, $ c $. Now this is already an acceptable answer, but I can rewrite certain things here slightly based on exponential and log properties, and this may be something that's expected of you. So let's go ahead and rewrite this using our exponential and log properties.

Now the thing that I see here is that I have this fraction raised to the power of $ x $ and I also have the natural log of a fraction. Now I can do some rewriting here, so $ \left(\frac{1}{2}\right)^x $. I can raise one and two in my numerator and denominator to the power of $ x $. So this gives me three times one to the power of $ x $ over two to the power of $ x $. Then when rewriting this natural log of one half, I could rewrite this as natural log of two to the power of negative one.

Then keeping all of my other terms the same here, $ 8^x $ over the natural log of eight plus that constant of integration $ c $. Now simplifying this to its most simple form, one to the $ x $ is just always going to be one. So my numerator here is just three times one, and that's just three. Then I'm going to move that two to the $ x $ to the bottom there. Then the natural log of two to the power of negative one, I can pull that negative one out front using my log property here to make this whole term negative.

So this becomes negative three over two to the power of $ x $ times the natural log of two. Then adding on my other term here, keeping it the same, $ 8^x $ over the natural log of eight plus that constant of integration $ c $. So now we've fully simplified this using our exponential and log properties, and this is our final answer. Now we're going to continue getting practice with this new rule coming up in the next couple of videos. I'll see you there.

Problem

Evaluate the indefinite integral.
$\int_{}^{} - {(6)}^{x} d x$

$- x 6^{x - 1} + C$

$- \frac{1}{x + 1} 6^{x + 1} + C$

$- \frac{{(6)}^{x}}{\ln 6} + C$

$- \frac{{(6)}^{x + 1}}{\ln x} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} 3 x^{4} - {(5)}^{x} d x$

$\frac{3}{5} x^{5} - \frac{5^{x + 1}}{\ln 5} + C$

$\frac{3}{5} x^{5} - \frac{5^{x}}{\ln 5} + C$

$12 x^{5} - \frac{5^{x}}{\ln 5} + C$

$12 x^{5} - \frac{5^{x + 1}}{\ln 5} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} 2 {(\frac{1}{3})}^{x} d x$

$- \frac{2}{3^{x} \ln 3} + C$

$- \frac{2 x^{x + 1}}{4} + C$

$- \frac{2 x}{\ln 3} + C$

$- \frac{2 x^{x + 1}}{3^{x} \ln 4} + C$

example

Integrals of General Exponential Functions Example 1

Video duration:

Video transcript

Here, we're asked to evaluate the integrals using a substitution. The first integral we have is the integral of four to the power of 7x dx. This is an exponential function, but instead of just x in the exponent, we have 7x, a function of x. Hence, we need to use a substitution.

When faced with a substitution like this one, typically, we'll want to set $ u $ equal to whatever is in the exponent, which is $ 7x $ here. If $ u = 7x $, then $ du $ equals the derivative of $ 7x $, which is $ 7 \, dx $. However, since I don't already have this seven in my integral, I need to multiply by the missing constant and its reciprocal to make this substitution work.

I multiply 4 to the power of 7x by seven and then include the dx. If I multiply by seven, I also need to multiply by $ \frac{1}{7} $ so that I'm not actually changing the value of my integral. Now I have $ u $ and $ du $ in the integral, which allows me to rewrite it in terms of $ u $ and $ du $ as $ \frac{1}{7} \times \int 4^u \, du $.

Then, I apply the rule for general exponential functions, yielding $ \frac{1}{7} \times \frac{4^u}{\ln(4)} $. Knowing $ u $, I replace it, so I get $ \frac{1}{7} \times \frac{4^{7x}}{\ln(4)} $ and I add the constant of integration $ C $ at the end, giving the final answer.

Let's take a look at another example, where we are asked to integrate $ 5^{\cos(\theta)} \sin(\theta) \, d\theta $. Although trig functions might seem concerning, we shouldn't worry, as this follows a similar approach. For $ 5^{\cos(\theta)} $, I set $ u $ equal to $ \cos(\theta) $. Given $ u = \cos(\theta) $, $ du $ equals the derivative of $ \cos(\theta) $, so $ -\sin(\theta) \, d\theta $.

We almost have $ du $ ready, we just need to adjust by a negative, so I change it to $ 5^{\cos(\theta)} (-\sin(\theta)) \, d\theta $ and adjust the integral accordingly. The revised integral is $ -\int 5^u \, du $.

Using the rule for general exponential functions: this gives $ -\frac{5^u}{\ln(5)} + C $. Plugging in $ u $, it simplifies to $ -\frac{5^{\cos(\theta)}}{\ln(5)} + C $, and this is the final answer. Here again, we used u-substitution along with the rule for general exponential functions.

We will get more practice with substitution and our exponential rule coming up next. I'll see you there.

Problem

Evaluate the definite integral.
$\int_{0}^{5} \frac{3^{\sqrt{x + 4}}}{\sqrt{x + 4}} d x$

49.15

73.73

77.91

32.77

Problem

Find the area under the graph of $f (x) = 3^{2 x}$ between $x equals 0$ and $x equals 2$ .

36.41

12.29

36.86

12.98

concept

Integrals of Natural Exponential Functions (e^x)

Video duration:

Video transcript

In the previous video, we learned how to find the integral of our general exponential function b to the power of x. Now, a common exponential function that we'll encounter is e to the power of x. So how do we then find this integral? Well, luckily, since e to the x is really just a special case of b to the x where our base is just e, we can use the integral rule that we just learned for our general exponential functions to now find the integral of e to the x. So that's exactly what we're going to do here.

Let's go ahead and jump right into things. Now looking at this integral exdx, we want to look to our rule for b to the x. And since our base here is e, we just want to replace all of these instances of b with e. So taking this interval is then going to give me exlne. Now looking at this, I know that the natural log of e is really just one.

So ex1 just gives me ex. So I'm left here with ex+c since we're dealing with an indefinite integral. So integrating e to the x just gives me e to the x. I'm left with the same function that I started with. Now this makes sense because we saw the same thing when taking the derivative of e to the x.

So what does this integral rule look like in action? Well, here we're faced with the integral of five times e to the x dx. Now five is just a constant. So I know using the constant multiple rule, I can just pull it out front, making this five times the integral of e to the x dx. Now based on our rule here, we will know that the integral of e to the x dx is just ex.

So this ultimately gives me five times ex+c since this is an indefinite integral. So now that we have this rule, we can use this along with all of our other integral rules that we've learned so far to find the integral of some more complicated functions that may now involve e. So looking at this integral here, we want to integrate 3x4−πex+2dx. Now depending on how comfortable you are with integrals at this point, you can do this all in one go. But I'm going to go ahead and break this integral down just so you can see what I'm doing step by step.

The first thing I'm going to do here is break this into three separate integrals using my sum and difference rules. So this becomes the integral of three x to the power of four dx minus the integral of pi times e to the x dx, and then plus the integral of two dx. So now we can focus on each of these integrals separately. So focusing on this first one here, the integral of three times x to the power of four dx. Three is just a constant.

So using my constant multiple rule, I know that I can pull this three out front. And just look at the integral of x to the power of four dx, which using the general power rule is going to be one fifth x to the power of five. So with that three, this is going to be three fifths x to the power of five. Then looking at our next integral here, pi times e to the x dx. Now remember that pi is a constant.

It is not a variable. So I can use my constant multiple rule here as well. Pulling that pi out front, focusing just on integrating e to the power of x dx. And based on the rule that we just learned, we know that that just leaves us with e to the x. So this is just pi times e to the x, the same function that we started with.

Then our final integral here, two dx two is just a constant. So this is going to give me two x, and then, of course, plus that constant of integration c on the end of my integral here. So this is my final answer, having applied multiple different rules for integration. Now we're going to continue getting practice with taking integrals and all of the rules that we've learned so far coming up next. I'll see you there.

Problem

Evaluate the indefinite integral.
$\int_{}^{} (\sqrt{x} - e^{x}) d x$

$\frac{2}{3} x^{\frac{2}{3}} - e^{x} + C$

$\frac{2}{3} x^{\frac{3}{2}} - x e^{x} + C$

$\frac{2}{3} x^{\frac{3}{2}} - e^{x} + C$

$\frac{2}{3} x^{\frac{2}{3}} - \frac{1}{x} e^{x} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} (4 e^{x} + \frac{1}{x^{3}}) d x$

$4 e^{x} - \frac{1}{4 x^{4}} + C$

$4 x e^{x} - \frac{1}{2 x^{2}} + C$

$4 x e^{x} - \frac{1}{4 x^{4}} + C$

$4 e^{x} - \frac{1}{2 x^{2}} + C$

example

Integrals of Natural Exponential Functions (e^x) Example 2

Video duration:

Video transcript

Here, we're asked to evaluate the indefinite integral using substitution. And the integral that we have here is 3xx2ex3 dx. Now upon first glance, this might look like a rather complicated integral. But using what we know about substitution, this is actually going to shake out to be a fairly straightforward one. So let's go ahead and get started here.

Now whenever we have an exponential function that has some function in its power, that's typically what we should choose to be u when doing substitution. So here, if I set u equal to x3, that then makes du equal to the derivative of x3, so 3x2dx. Now looking at my integral here, if this x3 is u, then I have du 3xx2 just nicely appear in my integral. So this allows me to rewrite this in terms of u and du rather easily because that 3xx2dx is du. This can then be rewritten as eudu.

Now we know that the integral of exdx is just ex and the same thing is true of eudu. So this is just going to be eu plus a constant of integration c. Now we know what u is, so we can go ahead and replace that with x3 there to give us our final answer. Now whenever we're working with a substitution that has this eudu, it's still always going to shake out to be eu, and then, of course, plus c if we're working with an indefinite integral. Now we're going to continue getting practice with this coming up in the next couple of practice problems.

I'll see you there.

Problem

Evaluate the indefinite integral.
$\int_{}^{} \frac{e^{2 \sqrt{x}}}{\sqrt{x}} d x$

$\frac{5 e^{2 \sqrt{x}}}{3 x^{\frac{3}{2}}} + C$

$\frac{3 e^{x^{\frac{7}{2}}}}{7 x^{\frac{3}{2}}} + C$

$e^{2 \sqrt{x}} + C$

$\frac{7 e^{x^{\frac{7}{2}}}}{3 x^{\frac{3}{2}}} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} e^{\sec θ} \sec θ \tan θ d θ$

$- \frac{e^{\sec θ}}{c o s^{2} ⁡ θ} + C$

$e^{\sec θ} + C$

$- e^{\sec θ} + C$

$\frac{e^{\sec θ}}{c o s^{2} ⁡ θ} + C$

example

Integrals of Natural Exponential Functions (e^x) Example 3

Video duration:

Video transcript

Here we're asked to evaluate the integrals using substitution. Now looking at this first integral that we're faced with evaluating here, it is et1 + et4 over et2 dt. Now looking at this array, the first thing I notice is that I have these e's raised to these powers that are actually functions. Now typically when doing a substitution, we often want to choose u to be that function that's in that exponent. But here, because we have two different functions happening in two different exponential functions, that's really not going to do anything for us.

So how else can we approach this problem? Well, looking at this here, since I only have one term in my denominator and I have two terms in that numerator, I can split this up into two separate fractions that I can then deal with separately. So let's see what happens when we do this. I'm splitting this up into et1 over et2 plus et4 over et2 dt. Now looking at this, I can simplify this further using exponential properties because I can go ahead and just subtract those powers because we're being divided and these exponential functions have the same base.

So this first term, et-2, that ends up giving me et-1, then doing the same thing to this next term here, et4. If I subtract four t minus two t, that's effectively what that bottom term is doing there, that leaves me with et2 and still with that dt. So now I can see that I can use a substitution on each of these two terms separately in order to ultimately get my answer here. So I'm going to go ahead and show you how I'm splitting this up. This is really the integral of et-1 dt plus the integral of et2 dt.

And we're effectively doing two different substitutions. Now for this first one here, I have et-1. So in doing a substitution here, if I make u equal to negative t, that makes du equal to negative dt. Now we don't have a negative there, so that means that we need to multiply by a negative on the inside, and in doing that we also have to multiply by a negative on the outside. So this is really giving us a negative integral of eu1 du.

We know that the integral of eu1 is just eu1 + c. So I have that negative tapped on the outside there. So this is negative eu1, and then we're going to worry about that constant later because we have one more integral to take. Now I know what this u is, so I'm going to go ahead and plug it back in to keep stuff organized. This is negative et-1.

So that's my antiderivative for this first term here. Let's take a look at this next one where we're going to do a separate substitution. Now for this one, we have our substitution that u is equal to two t. Now if u is equal to two t, that makes du equal to two dt. Now I don't have that two, so I need to go ahead and multiply by two.

And multiplying by two, I need to multiply by one-half as well so that I'm not actually changing the value of my integral. So this ends up being one-half integral of eu1 du. Now with this substitution, this gives me a one-half eu1, and then now I'm going to add on that constant of integration c. Now, again, here we want to make sure that we plug in what u actually was. It was two t, just keeping that organized.

And then this gives me my final answer here, negative et-1, plus one-half et2 + c. Now let's take a look at our second integral here. We're tasked with integrating three ex3, times one plus ex3 to the power of five dx. Now this does look like a rather complicated integral, but let's break this down. Now again, when working with substitutions, your first instinct is probably to make u equal to this three x.

But looking at this integral, seeing as it's more complicated, there's a bit more going on, that is not what we should do here. But we see that we have this ex3 outside of our parentheses as well. Now, going back to the basics of substitution, remember that we often wanted to choose something that was raised to a power. Now, here we can see that we have this whole thing here, one plus ex3 raised to the power of five. So what if we instead here make u equal to one plus ex3?

That makes du equal to the derivative of that. So the derivative of one is zero. I don't have to worry about it. And the derivative of ex3 using the chain rule is going to be three ex3, then of course multiplied by dx. So looking back at my integral here, if this is u, that makes du equal to three ex3 dx, which is exactly what we have left over in our integral.

So this then allows us to rewrite this as the integral of u to the power of five du. Now if I go ahead and do this integral, I just have to use the power rule. This gives me here one-sixth, u to the power of six plus c. Now, of course, we know what u is. It's that one plus ex3.

So we want to go ahead and plug that back in, making this one-sixth times one plus ex3 to the power of six plus c, and that's our final answer. So looking at these problems, we can see that even though oftentimes we are going to make u whatever function is in that exponent, that will not always be the case. And sometimes we have to get a bit creative in how we approach these problems. Now we're going to keep practicing these skills coming up next. I'll see you there.

Problem

Find the definite integral.
$\int_{- 1}^{1} x^{2} e^{x^{3}} d x$

0.78

0.86

0.31

-0.74

Problem

Find the area of $f (x) = 3 + e^{2 x}$ from $x equals 0$ to $x equals 2$ .

-20.80

32.80

-21.80

-21.30

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We have more practice problems on Integrals of Exponential Functions

Here’s what students ask on this topic:

How do you find the integral of an exponential function b^x?

To find the integral of an exponential function b^x, you reverse the derivative process. The integral is given by the formula:

\frac{b^{x}}{\ln (b)} + c

where b is the base of the exponential function, ln(b) is the natural logarithm of b, and c is the constant of integration. This formula is applicable when b is greater than zero and not equal to one. Understanding this rule allows you to integrate more complex functions involving exponential terms and constants.

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What is the integral of e^x?

The integral of e^x is a special case of the general exponential function integral. Since the base is e, the integral simplifies to:

e^{x} + c

where c is the constant of integration. This result makes sense because the derivative of e^x is also e^x. Therefore, when integrating e^x, you end up with the same function plus the constant of integration. This property is useful in solving more complex integrals involving e^x and other terms.

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How do you apply the constant multiple rule in integrals?

The constant multiple rule in integrals allows you to factor out constants from the integral. If you have an integral of the form:

\int k f (x) dx

where k is a constant, you can rewrite it as:

k \int f (x) dx

This simplifies the integration process by allowing you to focus on integrating the function f(x) separately. This rule is particularly useful when dealing with integrals involving exponential functions and other terms.

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What are the restrictions on the base b for the integral of b^x?

When finding the integral of b^x, there are certain restrictions on the base b. Specifically, b must be greater than zero and not equal to one. These restrictions are based on the definition of a general exponential function. If b were equal to one, the function would be constant, and the integral would not follow the standard exponential rules. Similarly, if b were less than or equal to zero, the function would not be defined in the real number system. These restrictions ensure the function behaves as expected in calculus.

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How do you integrate a function involving multiple exponential terms?

To integrate a function involving multiple exponential terms, you can use the sum rule to split the integral into separate parts. For example, if you have:

\int (a b^{x} + c^{x}) dx

you can rewrite it as:

a \int b^{x} dx + \int c^{x} dx

Then, apply the exponential integral rule to each term separately. This approach allows you to handle each exponential term individually, making the integration process more manageable.

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Integrals of Exponential Functions: Videos & Practice Problems

Integrals of General Exponential Functions

Video transcript

Evaluate the indefinite integral. ∫−(6)xdx\int_{}^{}-\left(6\right)^{x}dx

Evaluate the indefinite integral.∫3x4−(5)xdx\int_{}^{}3x^4-\left(5\right)^{x}dx

Evaluate the indefinite integral.∫2(13)xdx\int_{}^{}2\left(\frac13\right)^{x}dx

Integrals of General Exponential Functions Example 1

Video transcript

Evaluate the definite integral.∫053x+4x+4dx\int_0^5\frac{3^{\sqrt{x+4}}}{\sqrt{x+4}}dx

Find the area under the graph of f(x)=32xf\left(x\right)=3^{2x} between x=0x=0 and x=2x=2.

Integrals of Natural Exponential Functions (e^x)

Video transcript

Evaluate the indefinite integral.∫(x−ex)dx\int_{}^{}\left(\sqrt{x}-e^{x}\right)dx

Evaluate the indefinite integral.∫(4ex+1x3)dx\int_{}^{}\left(4e^{x}+\frac{1}{x^3}\right)dx

Integrals of Natural Exponential Functions (e^x) Example 2

Video transcript

Evaluate the indefinite integral.∫e2xxdx\int_{}^{}\frac{e^{2\sqrt{x}}}{\sqrt{x}}dx

Evaluate the indefinite integral.∫esec⁡θsec⁡θtan⁡θdθ\int_{}^{}e^{\sec\theta}\sec\theta\tan\theta d\theta

Integrals of Natural Exponential Functions (e^x) Example 3

Video transcript

Find the definite integral.∫−11x2ex3dx\int_{-1}^1x^2e^{x^3}dx

Find the area of f(x)=3+e2xf\left(x\right)=3+e^{2x} from x=0x=0 to x=2x=2.

Do you want more practice?

Here’s what students ask on this topic:

How do you find the integral of an exponential function b^x?

What is the integral of e^x?

How do you apply the constant multiple rule in integrals?

What are the restrictions on the base b for the integral of b^x?

How do you integrate a function involving multiple exponential terms?

Your Calculus tutors

Evaluate the indefinite integral.
$\int_{}^{} - {(6)}^{x} d x$

Evaluate the indefinite integral.
$\int_{}^{} 3 x^{4} - {(5)}^{x} d x$

Evaluate the indefinite integral.
$\int_{}^{} 2 {(\frac{1}{3})}^{x} d x$

Evaluate the definite integral.
$\int_{0}^{5} \frac{3^{\sqrt{x + 4}}}{\sqrt{x + 4}} d x$

Find the area under the graph of $f (x) = 3^{2 x}$ between $x equals 0$ and $x equals 2$ .

Evaluate the indefinite integral.
$\int_{}^{} (\sqrt{x} - e^{x}) d x$

Evaluate the indefinite integral.
$\int_{}^{} (4 e^{x} + \frac{1}{x^{3}}) d x$

Evaluate the indefinite integral.
$\int_{}^{} \frac{e^{2 \sqrt{x}}}{\sqrt{x}} d x$

Evaluate the indefinite integral.
$\int_{}^{} e^{\sec θ} \sec θ \tan θ d θ$

Find the definite integral.
$\int_{- 1}^{1} x^{2} e^{x^{3}} d x$

Find the area of $f (x) = 3 + e^{2 x}$ from $x equals 0$ to $x equals 2$ .