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

we're asked to find the definite integral from negative one to one of x squared times e to the x cubed dx. Now looking at this integral, I know that I'm going to have to do a substitution. So I'm going to start by treating this as an indefinite integral and then I'm going to apply my original bounds at the end. Now there are multiple ways that you could choose to evaluate an integral using substitution and either of those methods should still give you the same answer. So whatever you choose to do, you should still get the same answer as me. So let's get into things here. This is the integral of x squared times e to the x cubed dx. Now I know that I need to make a substitution where u is equal to that x cubed, so if u is equal to x cubed, that makes du equal to three x squared dx. Now in our integral here we have x squared dx but we're missing that three so we need to go ahead and multiply it by it. Now we also need to multiply by its reciprocal one third so that we're not actually changing the value of our integral. So now I can rewrite my integral as one third the integral of e to the u d u. This then gives me one third times e to the u plus c. Now I wanna go ahead and replace u with what it actually is here. It's x cubed. And then I don't actually have to worry about that constant c because this is bounded from negative one to one, and I'm gonna apply those original bounds now since this is back in as a function of x. So keeping that one third out front here, I'm plugging in that upper bound first, e to the power of one cubed and then minus e to the power of negative one cubed. Now one cubed is just going to be one and negative one cubed is just going to be negative one. So I'm going to go ahead and erase those off of here. Then this is one third times e because that's all e to the power of one is minus one over e since that's to the power of negative one. Now we could choose to simplify this down based on the fact that this is a fraction and I could give this a common denominator by multiplying by e over e. So this then gives me e squared minus one over e and I can go ahead and stick that one third here by putting that three in my denominator. So this gives me e squared minus one over three. Now this is an appropriate and correct answer depending on how you're expected to report your final answers. If we're not using a calculator, this is our final answer. But if we are, we can type this in and get a numerical answer in decimals. So I'm gonna go ahead and round to two decimal places here and and this will ultimately give me a 0.78 for my final answer. Now let us know if you have any questions in the comments. I'll see you in the next one.

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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

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

0.78

0.86

0.31

-0.74

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Verified step by step guidance

Recognize that the integral is a definite integral of the form ∫_{-1}^1 x^2 e^{x^3} dx. The limits of integration are -1 and 1.

Observe that the integrand, x^2 e^{x^3}, is an odd function multiplied by an even function. Specifically, x^2 is even (symmetric about the y-axis), and e^{x^3} is neither even nor odd, but the product x^2 e^{x^3} is even because x^2 dominates the symmetry.

Since the integrand is even, the integral from -1 to 1 can be simplified by doubling the integral from 0 to 1. This is because the area under the curve from -1 to 0 is symmetric to the area from 0 to 1. Rewrite the integral as 2∫_{0}^1 x^2 e^{x^3} dx.

To evaluate the integral, consider a substitution. Let u = x^3, which implies du = 3x^2 dx. Solve for x^2 dx in terms of du: x^2 dx = du/3.

Change the limits of integration according to the substitution. When x = 0, u = 0^3 = 0. When x = 1, u = 1^3 = 1. The integral becomes 2∫_{0}^1 (1/3)e^u du. Factor out the constant 1/3, leaving (2/3)∫_{0}^1 e^u du, which can be evaluated using the antiderivative of e^u.