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We will rewrite that as du = 6r? dr. Our goal is now to rewrite the entire integral in terms of our new variable u and its derivative du. Since du = 6x2 dx and our integrand includes r? dr, we will write - du = x² dx. So we can rewrite the integral as follows: |7(2° + 3)° dr = du. This new integral is one that we can do easily. Complete this integral, getting an answer in terms of u.

We will rewrite that as du = 6r? dr. Our goal is now to rewrite the entire integral in terms of our new variable u and its derivative du. Since du = 6x2 dx and our integrand includes r? dr, we will write - du = x² dx. So we can rewrite the integral as follows: |7(2° + 3)° dr = du. This new integral is one that we can do easily. Complete this integral, getting an answer in terms of u.

Calculus: Early Transcendentals
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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5. We will rewrite that as \( du = 6x^2 \, dx \). Our goal is now to rewrite the entire integral in terms of our new variable \( u \) and its derivative \( du \). Since \( du = 6x^2 \, dx \) and our integrand includes \( x^2 \, dx \), we will write \( \frac{1}{6} \, du = x^2 \, dx \). So we can rewrite the integral as follows: \[ \int x^2 (2x^3 + 3)^3 \, dx = \int \frac{1}{6} u^3 \, du. \] This new integral is one that we can do easily. Complete this integral, getting an answer in terms of \( u \).
Transcribed Image Text:5. We will rewrite that as \( du = 6x^2 \, dx \). Our goal is now to rewrite the entire integral in terms of our new variable \( u \) and its derivative \( du \). Since \( du = 6x^2 \, dx \) and our integrand includes \( x^2 \, dx \), we will write \( \frac{1}{6} \, du = x^2 \, dx \). So we can rewrite the integral as follows: \[ \int x^2 (2x^3 + 3)^3 \, dx = \int \frac{1}{6} u^3 \, du. \] This new integral is one that we can do easily. Complete this integral, getting an answer in terms of \( u \).
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