Question

In Exercises \(1-10\), evaluate the integral. $$ \int_{0}^{x^{3}} y e^{-y / x} d y $$

Short answer

The evaluated integral is \( -x^{4} + x \)

Step-by-step solution

Integrate the Integral

To find the antiderivative of the integrand in \( \int_{0}^{x^{3}} y e^{-y / x} d y \), consider \( y e^{-y / x} \) as \( f(y) = y \times g(y) \) where \( g(y) = e^{-y/x} \). Apply integration by parts, given by the formula \( \int u dv = uv - \int v du \), with \( u = y \) and \( dv = e^{-y/x} dy \). The derivatives will be \( du = dy \) and \( v = -x e^{-y / x} \). Substituting these into the formula will yield the integral as \( -y x e^{-y / x} - \int -x e^{-y/x} dy \), which simplifies to \( -xy e^{-y / x} + x \int e^{-y/x} dy \). Next, consider the integral \( \int e^{-y/x} dy \) and substitute \( z = -y/x \), then \( dy = -x dz \). This simplifies the integral as \( x \int e^{z} dz \), which is equal to \( xe^{z} + C \), or \( -x e^{-y / x} + C \). Therefore, the whole integral simplifies to \( -xy e^{-y / x} - x e^{-y / x} + C = -x( y + 1) e^{-y / x} + C \).

Evaluate the Definite Integral

Plug in the limits of the integral \( \int_{0}^{x^{3}} -x( y + 1) e^{-y / x} dy \). First, substitute the upper limit \( y = x^{3} \) into the antiderivative. Then plug in the lower limit \( y = 0 \) into the antiderivative and subtract the former from the latter according to the Fundamental Theorem of Calculus \( F(b) - F(a) \). The evaluated integral will be \( -x( x^3 + 1 ) e^{-x^{3} / x} + x( 0 + 1 ) e^{-0 / x} = -x^{4} + x \).

Simplify the Result

Simplify the result from the previous step to get the final answer. The resulting value of the integral is \( -x^{4} + x \).