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Find the area of the region bounded by the given curves.

$ y = x^2 e^{-x} $ , $ y = xe^{-x} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Missouri State University

Campbell University

Baylor University

University of Nottingham

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

02:01

Find the area of the regio…

04:14

The problem is find this area of the region bounded by given curves y is equal to x, squared times to negative x and y is equal to x times e to negative x point so first, we like these 2 functions, are equal and find the intersection point. So this x, squared minus x times into negative x, is equal to 0 point. So we have x is equal to 0 or x is equal to 1 o the interval is 0 to 1 and when x is between these 2 points we have x times e to negative x is greater than x squared times e to negative x cos ever Of this region is integral, from 1 to 2 x, minus x, squared to negative now for this problem. We can use integration by parts. The formula is the integral of: u times 3 x is equal to: u times, v minus the integral of u prim times. V d: we cannot? U is equal to x, minus x, squared and promise to negative x. Then you primis, equal to 1 minus 2 x and v, is equal to negative 2 negative x, and this integral is equal to by this formula. This is equal to negative x, minus x, squared times e to negative x from 1 to 2 and minus integral of prime times, so this is as integral from 1 to 21 minus 2 x e to negative x dx. So this is equal to it. For the first term, so this is negative were plugging 2 and 1 is its function. That is negative, 2 minus 4 into negative 2 and minus 0, and then this integral. We can also use u substitution and that, u is equal to 1 minus 2 x and prime, is equal to 2 negative x, so prim is equal to negative 2 and v is equal to negative into negative, like potent integral is equal to u times so. This is 1 minus 2 x times negative: u to negative x from 1 to 2 and minus integral of your prime times. So this is 2 to negative from 1 to 2, and now is is from 0 to 1 o sorry. This is from 0 to 1021 o 021 and then this value is 0 minus 0 point, so the first term is 0 and there. The second term is 1 minus 2 x times negative into negative x from 0 to 1 minus the integral from 0 to 1. You cram have sped is 2 times e to tie. This is equal to pay 10 to this function. We have. This is negative 1 times negative e to negative 1 is to negative 1 minus. This is 1 times negative 1. This is possible and then the is minus 10 in negative 2 to negative 0 to 1. This is equal to 2 negative 1 plus 11 minus negative 2 times 2 to 1 and 2 is. The answer is 3 times e to the negative 1 and minus 1.

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