Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

0π2sinx(1+cosx)dx

Short Answer

Expert verified

0π2sinx(1+cosx)dx=32.

Step by step solution

01

Step 1. Given information.

A definite integral is given as0π2sinx(1+cosx)dx.

02

Step 2. Using the Fundamental theorem of Calculus.

We get

0π2sinx(1+cosx)dx=0π2sinxdx+0π2sinxcosxdx=[-cosx]0π2+120π22sinxcosxdx=-cosπ2+cos0+120π2sin(2x)dx=-0+1+12[-12cos(2x)]0π2=1-14[cosπ-cos0]=1-14(-1-1)=1+12=32

So the exact value of the given definite integral is32.

03

Step 3. The graph to verify the answer is

The solution is area under graph which is

a=1.5=32

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free