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Chapter 5: Problem 43

Use a change of variables to evaluate the following definite integrals. $$\int_{-1}^{2} x^{2} e^{x^{3}+1} d x$$

Short Answer

Expert verified
Answer: The value of the definite integral is $(8)^{\frac{1}{3}}e^9 - 1$.

Step by step solution

01

Apply Substitution

Let's perform the substitution: \(u = x^3 + 1\). Now find the derivative of \(u\) with respect to \(x\), and solve for \(dx\): $$\frac{d u}{d x} = 3x^2$$ $$dx = \frac{du}{3x^2}$$ We also need to change the integration limits according to the new variable, \(u\): When \(x=-1\), \(u=(-1)^3 + 1 = 0\) When \(x=2\), \(u=(2)^3 + 1 = 9\) Thus, the integral transforms into: $$\int_{0}^{9} e^u \frac{du}{3x^2}$$ Notice that we still have \(x^2\) in the denominator. We can solve for \(x^2\) from the substitution \(u = x^3 + 1\): $$x^3 = u-1$$ $$x^2 = (\sqrt[3]{u-1})^2$$ Now the integral becomes: $$\int_{0}^{9} e^u \frac{du}{3 (\sqrt[3]{u-1})^2}$$
02

Evaluate the Integral

Evaluate the resulting integral: $$\int_{0}^{9} e^u \frac{du}{3 (\sqrt[3]{u-1})^2}$$ Let's pull the constant factor 1/3 out of the integral: $$\frac{1}{3}\int_{0}^{9} e^u\frac{du}{(\sqrt[3]{u-1})^2}$$ Now, we can use the power rule for integration: $$\frac{1}{3}\int e^u (u-1)^{-\frac{2}{3}} du$$ Integrating with respect to \(u\), we get: $$\frac{1}{3}(3(u-1)^{\frac{1}{3}}e^u)|_0^9$$
03

Evaluate the Definite Integral

Now, we will plug in the new limits of integration and find the value of the integral: $$\frac{1}{3}(3(u-1)^{\frac{1}{3}}e^u)|_0^9 = (u-1)^{\frac{1}{3}}e^u|_0^9$$ Evaluate the definite integral at the upper and lower limits: $$[(9-1)^{\frac{1}{3}}e^9 - (0-0)^{\frac{1}{3}}e^0]$$ $$= (8)^{\frac{1}{3}}e^9 - 1$$ Hence, the value of the definite integral is: $$\int_{-1}^{2} x^{2} e^{x^{3}+1} d x = (8)^{\frac{1}{3}}e^9 - 1$$

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