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$\int_{0}^{1} x^{5} e^{1-x^{6}} \, dx$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int_{0}^{1} x^{5} e^{1-x^{6}} \, dx

Identify Integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $x^5 * e^{1-x^6}$ from $0$ to $1$ .
Substitution Simplification: Look for a substitution that simplifies the integral. $\newline$ Let $u = 1 - x^6$ , which implies that $du = -6x^5 dx$ . We can solve for $dx$ to get $dx = -\frac{du}{6x^5}$ .
Change Limits: Change the limits of integration according to the substitution. $\newline$ When $x = 0$ , $u = 1 - 0^6 = 1$ . $\newline$ When $x = 1$ , $u = 1 - 1^6 = 0$ . $\newline$ So the new limits of integration are from $u = 1$ to $u = 0$ .
Rewrite in Terms of $u$ : Rewrite the integral in terms of $u$ . $\newline$ The integral becomes: $\newline$ $\int_{1}^{0} x^5 \cdot e^u \cdot \left(-\frac{du}{6x^5}\right)$ $\newline$ The $x^5$ terms cancel out, and we can pull out the constant factor of $-\frac{1}{6}$ : $\newline$ $-\frac{1}{6} \cdot \int_{1}^{0} e^u \, du$
Evaluate Integral: Evaluate the integral with the new limits. $\newline$ The integral of $e^u$ with respect to $u$ is $e^u$ , so we have: $\newline$ $-\frac{1}{6} \cdot [e^u]_{1}^{0}$
Apply Fundamental Theorem: Apply the Fundamental Theorem of Calculus. We substitute the limits of integration into the antiderivative: $-\frac{1}{6} * (e^{0} - e^{1})$
Simplify Expression: Simplify the expression. $\newline$ Since $e^0 = 1$ and $e^1 = e$ , we have: $\newline$ $-\frac{1}{6} \times (1 - e)$
Check for Errors: Check for any mathematical errors. $\newline$ There are no mathematical errors in the previous steps.

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