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Chapter 16: Problem 22

Evaluate the following iterated integrals. $$\int_{0}^{2} \int_{0}^{1} x^{5} y^{2} e^{x^{3} y^{3}} d y d x$$

Short Answer

Expert verified
Question: Evaluate the iterated integral $\int_{0}^{2}\int_{0}^{1} x^{5} y^{2} e^{x^{3} y^{3}} dydx$. Answer: $\frac{1}{3} \int_{0}^{2} e^{x^3} dx - \frac{2}{3}$

Step by step solution

01

Integrate with respect to y

First, we will integrate the inner integral with respect to \(y\). The integral we want to compute is: $$\int_{0}^{1} x^{5} y^{2} e^{x^{3} y^{3}} dy$$ To do this, we will use substitution method. Let \(u = x^3y^3\). Then: $$\frac{du}{dy} = 3x^3y^2$$ Now, we can rewrite the integral as: $$\int_{0}^{1} x^{5} y^{2} e^{x^{3} y^{3}} dy = \frac{1}{3} \int e^u du$$ Now, integrating e^u with respect to u: $$\frac{1}{3} \int e^u du = \frac{1}{3} e^u + C_1 = \frac{1}{3} e^{x^3y^3} + C_1$$
02

Apply the limits of integration for y

Now we will apply the limits of integration for \(y\): $$\Big[\frac{1}{3} e^{x^3y^3} \Big]_0^1 = \frac{1}{3} e^{x^3} - \frac{1}{3} e^0 = \frac{1}{3} e^{x^3} - \frac{1}{3}$$
03

Integrate with respect to x

Now we can integrate the outer integral with respect to \(x\). The integral we want to compute is: $$\int_{0}^{2} \left( \frac{1}{3} e^{x^3} - \frac{1}{3} \right) dx$$ This can be separated into two integrals: $$\frac{1}{3} \int_{0}^{2} e^{x^3} dx - \frac{1}{3} \int_{0}^{2} dx$$
04

Integrate e^x^3

There is no simple way to integrate \(e^{x^3}\) with respect to \(x\), so we'll denote this integral by \(I\). This step will be: $$I = \int_{0}^{2} e^{x^3} dx$$
05

Integrate the constant 1

Now we can integrate the constant 1 with respect to \(x\): $$\int_{0}^{2} dx = \Big[x\Big]_0^2 = 2 - 0 = 2$$
06

Combine the integrals and apply the limits of integration for x

Now, combining the integrals, we get: $$\frac{1}{3} \left( I - 2 \right) = \frac{1}{3} \int_{0}^{2} e^{x^3} dx - \frac{2}{3}$$ Since there is no simple way to find the integral of \(e^{x^3}\), the final answer will be an expression that involves the undefined integral I: $$\frac{1}{3} \int_{0}^{2} e^{x^3} dx - \frac{2}{3}$$

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