   Chapter 5.1, Problem 35E ### 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. ∫ 5 x + 4 x 3   d x

To determine

To calculate: The value of indefinite integral (5x+4x3)dx and check the result by differentiation.

Explanation

Given Information:

The indefinite integral is (5x+4x3)dx.

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 (5x+4x3)dx.

The integrand of the indefinite integral is (5x+4x3).

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

(5x+4x3)dx=(5xx3)dx+(4x3)dx=5(xx13)dx+4(1x13)dx=5(x113)dx+4(x13)dx=5(x23)dx+4(x13)dx

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

(5x+4x3)dx=5[x23+123+1]+4[x13+113+1]+C=5[x5353]+4[x2323]+C=5(35)x53+4(32)x23+C=3x35+6x23+C

Simplify further the above integration

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