Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 59 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 59 maths.

Answers (1)

Answer :   6

Hint: Use indefinite integral formula and the limits to solve this integral

Given: \int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx

Solution:

\begin{aligned} &\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x\\ &=\int_{\frac{1}{3}}^{1} \frac{x\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{4}} d x=\int_{\frac{1}{3}}^{1} \frac{\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{3}} d x\ \end{aligned}

Put \frac{1}{x^2}-1=t\Rightarrow \frac{-2}{x^3}dx=dt\Rightarrow dx=\frac{-x^3}{2}dt
When x=\frac{1}{3}  then t=8 and when x=1 then t=0

 

\begin{aligned} &\int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x=\int_{\frac{1}{3}}^{1} \frac{\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{3}} d x \\ &=-\int_{8}^{0} \frac{t^{\frac{1}{3}}}{2} d t \\ &=-\frac{1}{2} \int_{8}^{0} t^{\frac{1}{3}} d t=-\frac{1}{2}\left[\frac{t^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{8}^{0}=-\frac{1}{2}\left[\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]_{8}^{0} \\ &=-\frac{1}{2} \times \frac{3}{4}\left[t^{\frac{4}{3}}\right]_{8}^{0}=\frac{-3}{8}\left[0^{\frac{4}{3}}-8^{\frac{4}{3}}\right] \\ &=\frac{-3}{8}\left[0-2^{4}\right]=\frac{-3}{8}(-16)=2 \times 3=6 \end{aligned}

Posted by

infoexpert24

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

CBSE Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question