   Chapter 5.3, Problem 21E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Evaluate the integral. ∫ 0 2 ( 4 5 t 3 − 3 4 t 2 + 2 5 t )   d t

To determine

To evaluate: The integral function.

Explanation

Given:

The integral function is 02(45t334t2+25t)dt.

Calculation:

The integral function is 02(45t334t2+25t)dt (1)

Integrate Equation (1) with respect to t.

02(45t334t2+25t)dt=0245t3dt0234t2dt+0225tdt=(45×t44)02+(34×t33)02+(25×t22)02=(t45t34+t25)02 (2)

Apply the upper and lower limits in the equation (2)

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