   Chapter 5.1, Problem 34E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding indefinite integrals In Exercises 25–36, find the indefinite integral. Check your result by differentiating. See Examples 4 and 5. ∫ t 2 + 2 t 2   d t

To determine

To calculate: The value of indefinite integral (t2+2t2)dt and check the result by differentiation.

Explanation

Given Information:

The indefinite integral is (t2+2t2)dt.

Formula used:

The simple power rule of integration, xndx=xn+1n+1+C where, n1.

The sum or difference rule of integration, (f(x)±g(x))dx=f(x)dx±g(x)dx.

The simple power rule for the derivative, ddx[xn]=nxn1.

Calculation:

Consider the indefinite integral (t2+2t2)dt.

The integrand of the indefinite integral is (t2+2t2).

Integrate the provided indefinite integral use the sum or difference rule of integration, (f(x)±g(x))dx=f(x)dx±g(x)dx.

(t2+2t2)dt=t2t2dt+2t2dt=1dt+2t2dt

Integrate further use the simple power rule of integration, xndx=xn+1n+1+C

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