Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 49 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 49 Maths Textbook Solution.

Answers (1)

Answer:

\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx

Sol:

\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx=\int_{0}^{1} xe^{x}dx+\int_{0}^{1}\cos \frac{\pi x}{4}dx

Applying integration by parts method in !st integral then

=\left[x \int e^{x} d x\right]_{0}^{1}-\int_{0}^{1}\left(\frac{d(x)}{d x} \int e^{x} d x\right) d x+\left[\frac{\sin \frac{\pi}{4} x}{\frac{\pi}{4}}\right]_{0}^{1}

=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 . e^{x} d x+\frac{4}{\pi}\left[\sin \frac{\pi x}{4}\right]_{0}^{1}

=\left[1 . e^{1}-0 . e^{0}\right]-\left[e^{x}\right]_{0}^{1}+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]

=[e-0]-\left[e^{1}-e^{0}\right]+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]

= \left [ e \right ]-[e-1]+\frac{4}{\pi}\left[\frac{1}{\sqrt{2}}-0\right]

= e-e+1+\frac{4}{\pi} \times \frac{1}{\sqrt{2}}

=\frac{2\sqrt{2}}{\pi }+1

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

SC asks CBSE to protect students affected by OSM evaluation in admissions
anam-khan

CBSE Class 12 Compartment Result 2026 LIVE: 12th supplementary marksheet out at results.digilocker.gov.in
anam-khan

CBSE Class 12 supplementary results 2026 out; 53.08% students pass

Latest Question