10th Edition

Ron Larson + 1 other

ISBN: 9781285057095

Textbook Problem

Finding or Evaluating an Integral In Exercises 1–8, use the basic integration rules to find or evaluate the integral.

∫ x x 2 − 36 d x

To determine

To calculate: The value of the indefinite integral ∫xx2−36dx.

Solution:

The value of the indefinite integral ∫xx2−36dx is 13(x2−36)3/2+C.

The integral is ∫xx2−36dx.

Formula used:

The power formula,

∫undu=un+1n+1+C

Calculation:

Consider the integral y=∫xx2−36dx.

Rewrite the equation as,

y=∫x2−36(xdx)

Let u=x2−36.

Differentiate both sides with respect to x.

dudx=ddx(x2−36)dudx=ddx(x2)−ddx(36)dudx=2x+0du=2xdx

Substitute u for x2−36 and du2 for xdx.

y=∫u(du2)=12∫udu=12∫u1/2du

Solve the above integral as below,

y=12[u12+112+1]+C=12[u32(32)]+C=12[23u32]+C=13u32+C

Resubstitute x2−36 for u.

y=13(x2−36)3/2+C

Hence, the value of the indefinite integral ∫xx2−36dx is 13(x2−36)3/2+C.

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Calculus

Finding or Evaluating an Integral In Exercises 1–8, use the basic integration rules to find or evaluate the integral. ∫ x x 2 − 36 d x

The Solution to Your Study Problems

Chapter 8 Solutions

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Math
Calculus Calculus 10th Edition

Finding or Evaluating an Integral In Exercises 1–8, use the basic integration rules to find or evaluate the integral.

∫ x x 2 − 36 d x