Find the area under the graph of f(t)=t2et+tt2f\left(t\right)=\frac{t^2e^{t}+t}{t^2} from t=1t=1 to t=3t=3.

e 3 − e + ln ⁡ 3 ≈ 18.47 e^3-e+\ln3\thickapprox18.47

Find the area under the graph of f ( t ) = t 2 e t + t t 2 f\left(t\right)=\frac{t^2e^{t}+t}{t^2} from t = 1 t=1 to t = 3 t=3 .

Here we're asked to find the area under the graph of the function f of t equals t squared times e to the t plus t over t squared from t equals one to t equals three. Now remember that to find the area under a curve, we just need to set up an integral. So here, this is going to be equal to the integral from one to three of our function here, t squared e to the t plus t over t squared d t and evaluating this integral will give us our area. Now looking at this integral, since I have two terms in that numerator and only one term in the denominator, that's usually a good clue that I should just break this up into two separate fractions to simplify it a bit. So rewriting this as two separate fractions, the integral from one to three of t squared e to the t over t squared, and then plus t over t squared dt. Now we see that some stuff is going to cancel here. This t squared will cancel with that t squared and all we're left with there is e to the power of t, and then t over t squared, one of those t's will cancel on the bottom and I'm just left there with one over t. So this is really the integral from one to three of e to the t plus one over t dt and these are two much easier functions to find the antiderivative of. So I know the antiderivative of e to the t is just going to be that same function e to the t and then one over t here, that's going to result in the natural log of the absolute value of t and then this is bounded from one to three. So now we wanna go ahead and plug in those bounds. When we're plugging in our bounds here, these absolute value signs no longer matter because one and three are already positive values. So this is e cubed plus the natural log of three. That's my upper bound plugged in. Subtracting off my lower bound plugged in, e to the power of one, that's just e. So e plus the natural log of one. The natural log of one is just zero. So that whole term goes away here and we're just left with e cubed plus the natural log of three minus e. Now there's not really a way to simplify this much. I could rewrite my e's both together at least. So e cubed minus e plus the natural log of three, but this is ultimately my final answer here if I can't use a calculator. Now if you can use a calculator you can type this in to get a answer in terms of decimals and rounding that off into two decimal places this is going to end up being about 18.47 and and that's my final answer, the area under my curve from one to three. Let us know if you have any questions, and I'll see you in the next one.

Find the area under the graph of f ( t ) = t 2 e t + t t 2 f\left(t\right)=\frac{t^2e^{t}+t}{t^2} from t = 1 t=1 to t = 3 t=3 .

e 3 − e + ln ⁡ 3 ≈ 18.47 e^3-e+\ln3\thickapprox18.47

4 e 3 + ln ⁡ 3 − 2 e ≈ 76.00 4e^3+\ln3-2e\thickapprox76.00

4 e 3 + ln ⁡ 3 − 2 e − 1 ≈ 75.00 4e^3+\ln3-2e-1\thickapprox75.00

e 2 + ln ⁡ 3 ≈ 8.49 e^2+\ln3\thickapprox8.49

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

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Multiple Choice

Find the area under the graph of $f (t) = \frac{t^{2} e^{t} + t}{t^{2}}$ from $t equals 1$ to $t equals 3$ .

$4 e^{3} + \ln 3 - 2 e \approx 76.00$

$4 e^{3} + \ln 3 - 2 e - 1 \approx 75.00$

$e^{3} - e + \ln 3 \approx 18.47$

$e^{2} + \ln 3 \approx 8.49$

Verified step by step guidance

Step 1: Start by simplifying the given function f(t) = (t^2e^t + t) / t^2. Divide each term in the numerator by t^2 to simplify the expression. This results in f(t) = e^t + 1/t.

Step 2: To find the area under the curve from t = 1 to t = 3, set up the definite integral of f(t) with respect to t over the interval [1, 3]. The integral becomes ∫[1,3] (e^t + 1/t) dt.

Step 3: Break the integral into two parts for easier computation: ∫[1,3] e^t dt + ∫[1,3] (1/t) dt. This allows us to handle each term separately.

Step 4: Compute the first integral, ∫[1,3] e^t dt. The antiderivative of e^t is e^t, so evaluate it as e^t | from t = 1 to t = 3, which gives e^3 - e.

Step 5: Compute the second integral, ∫[1,3] (1/t) dt. The antiderivative of 1/t is ln|t|, so evaluate it as ln(t) | from t = 1 to t = 3, which gives ln(3) - ln(1). Combine the results from both integrals to get the final expression for the area under the curve.

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