Evaluate the definite integral. ∫ 0 5 3 x + 4 x + 4 d x \int_0^5\frac{3^{\sqrt{x+4}}}{\sqrt{x+4}}dx

we're asked to evaluate a definite integral. And here we have the definite integral from zero to five of three to the power of the square root of x plus four over the square root of x plus four d x. Now looking at this integral, it looks like there's quite a bit of stuff going on, but let's go ahead and look at this step by step. Now remember that when faced with a definite integral, when we have to do a substitution, as it looks like we will have to do here, there are two different ways you could go about evaluating it. I'm gonna go ahead and show one of these methods, which is just ignoring our balance until the very end and then applying our original balance. You could choose to change your balance along with your substitution if you please, and you should get the same answer. So I'm just gonna focus on my integrand for now. Three to the power of the square root of x plus four. Because this is an exponential function, I know that I'm probably gonna wanna set u equal to the square root of x plus four to do a substitution here and ultimately find my integral. So if u is equal to the square root of x plus four, what does that make du? Well, remember that the square root of x plus four is really the same thing as x plus four to the power of one half. So So if we go ahead and use the power rule here, this is going to give us a one half times x plus four to the power of negative one half. Now because using the chain rule here is just going to give me my derivative of one, one, multiplying by one doesn't change anything. So this is ultimately my derivative and then I wanna multiply this by DX. Now if I rewrite this slightly, this becomes one over two times one over the square root of X plus four, which is exactly denominator of my integrand here. So I have U and I have part of DU but I'm just missing this one half. So because I'm missing that one half, I can go ahead and multiply by one half. So this is one half times three times the square root of x plus four over the square root of x plus four DX. And then because I am multiplying by this missing constant, I also need to multiply by its reciprocal, which in this case is two, so that I'm not actually changing the value of my integral. So now I have u and I also have d u. So I can rewrite this as two times the integral of three to the power of u d u. Now this is just a basic general exponential function. So this is two times three to the power of u over the natural log of three, and then right now plus my constant of integration c. Now I can go ahead and change u to be what it originally was. It's the square root of x plus four, so putting that function of x back in there. Then I know that I actually am working with a definite integral, so I really don't need that plus c. And I can go ahead and bound this from zero to five. Now we can just plug in those bounds. So this is two over the natural log of three. I'm just gonna keep all my constants out front there times three to the power of five plus four, my upper bound plugged in, minus three to the power of the square root of zero plus four. Now looking at this here, keeping that two over the natural log of three out front, the square root of five plus four is really the square root of nine which is just three. So this is three cubed minus three to the power of the square root of zero plus four is really three squared. Doing some more simplification here, two over the natural log of three and this is times 27 minus nine. That's what three cubed and three squared are. 27 minus nine is going to give us here 18. So this is two times 18 over the natural log of three, or this is also just equal to 36 over three, just doing that multiplication or 36 over the natural log of three. Now this is an acceptable answer if you cannot use a calculator in your course, but if you can use a calculator and you're expected to fully get that decimal answer when you type this into your calculator, this will ultimately be equal to 32.77. And this is our final answer for that definite integral. If you have any questions, feel free to let us know in the comments, and let's keep practicing.

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

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Multiple Choice

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

Verified step by step guidance

Recognize that the integral involves the expression $ \frac{3^{\sqrt{x+4}}}{\sqrt{x+4}} $, which suggests a substitution method is appropriate. Let $ u = \sqrt{x+4} $, so $ u^2 = x+4 $. Differentiate both sides to get $ 2u \, du = dx $.

Adjust the limits of integration to match the substitution. When $ x = 0 $, $ u = \sqrt{0+4} = 2 $. When $ x = 5 $, $ u = \sqrt{5+4} = 3 $. The new limits of integration are from $ u = 2 $ to $ u = 3 $.

Substitute $ x+4 = u^2 $ and $ dx = 2u \, du $ into the integral. The integral becomes $ \int_2^3 3^u \, 2u \, \frac{1}{u} \, du $, which simplifies to $ 2 \int_2^3 3^u \, du $.

Recall the formula for the integral of an exponential function: $ \int a^u \, du = \frac{a^u}{\ln(a)} + C $, where $ a > 0 $ and $ a \neq 1 $. Apply this formula to $ \int 3^u \, du $, yielding $ \frac{3^u}{\ln(3)} $.

Evaluate the definite integral by substituting the limits $ u = 3 $ and $ u = 2 $ into $ \frac{3^u}{\ln(3)} $. The result is $ 2 \left[ \frac{3^3}{\ln(3)} - \frac{3^2}{\ln(3)} \right] $. Simplify this expression to find the final value.