copriate substitution to evaluate the integral [2√5 + 5x + 2 da.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem Statement**

Use an appropriate substitution to evaluate the integral:

\[
\int x \sqrt{5x + 2} \, dx.
\]

**Solution Approach**

To solve the integral, identify a substitution that simplifies the expression under the square root. Set \( u = 5x + 2 \), which implies \( du = 5 \, dx \) or \( dx = \frac{du}{5} \).

Next, express \( x \) in terms of \( u \):

\[ 5x + 2 = u \implies x = \frac{u - 2}{5}. \]

Now, substitute back into the integral:

\[
\int x \sqrt{5x + 2} \, dx = \int \frac{u - 2}{5} \sqrt{u} \cdot \frac{du}{5}.
\]

This substitution transforms the integral into a form that is easier to evaluate. Simplify and integrate to find the solution.

**Further Explanation**

Such problems help in understanding the application of the substitution method, a fundamental technique in calculus used to simplify and compute complex integrals.
Transcribed Image Text:**Problem Statement** Use an appropriate substitution to evaluate the integral: \[ \int x \sqrt{5x + 2} \, dx. \] **Solution Approach** To solve the integral, identify a substitution that simplifies the expression under the square root. Set \( u = 5x + 2 \), which implies \( du = 5 \, dx \) or \( dx = \frac{du}{5} \). Next, express \( x \) in terms of \( u \): \[ 5x + 2 = u \implies x = \frac{u - 2}{5}. \] Now, substitute back into the integral: \[ \int x \sqrt{5x + 2} \, dx = \int \frac{u - 2}{5} \sqrt{u} \cdot \frac{du}{5}. \] This substitution transforms the integral into a form that is easier to evaluate. Simplify and integrate to find the solution. **Further Explanation** Such problems help in understanding the application of the substitution method, a fundamental technique in calculus used to simplify and compute complex integrals.
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