Get Answers to all your Questions

header-bg qa

Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 33 textbook solution.

Answers (1)

Answer:-  \frac{16}{15}\sqrt{2}

Hints:-  You must know the rules of integration.

Given:-  \int_{0}^{2} x \sqrt{2-x} \cdot d x

Solution : I=\int_{0}^{2} x \sqrt{2-x} \cdot d x

   I=\int_{0}^{2}(2-x) \sqrt{x} \cdot d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]

\begin{aligned} &I=\int_{0}^{2} 2 \sqrt{x} d x-\int_{0}^{2} x^{3 / 2} x \cdot d x \\ &\left.\left.=2 \times \frac{2}{3} x^{3 / 2}\right]_{0}^{2}-\frac{2}{5} x^{5 / 2}\right]_{0}^{2} \\ &I=2 \sqrt{2}\left(\frac{4}{3}-\frac{4}{5}\right) \\ &I=\frac{16}{15} \sqrt{2} \end{aligned}

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

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