#### Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 1 Suquestion (ii) Maths Textbook Solution.

Answer:  $3-\frac{\pi}{2}+e^{6}$

Hint: Break the range of integration from $0 \: to \: \frac{\pi}{2},\: \frac{\pi}{2} \: to\: 3,$ and then $3\: to \: 9$

Given:

$\int_{0}^{9} f(x) d x \text { where } f(x)=\left\{\begin{array}{ll} \sin x, & \text { if } 0 \leq x \leq \frac{\pi}{2} \\\\ 1, & \text { if } \frac{\pi}{2} \leq x \leq 3 \\\\ e^{x-3}, & \text { if } 3 \leq x \leq 9 \end{array}\right\}$

Solution:

$\int_{0}^{9} f(x) d x$

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} f(x) d x+\int_{\frac{\pi}{2}}^{3} f(x) d x+\int_{3}^{9} f(x) d x \\ & \end{aligned}

$=\int_{0}^{\frac{\pi}{2}} \sin x d x+\int_{\frac{\pi}{2}}^{3} 1 d x+\int_{3}^{9} e^{x-3} d x \\$

$=[-\cos x]_{0}^{\frac{\pi}{2}}+[x]_{\frac{\pi}{2}}^{3}+\left[\frac{e^{x}}{e^{3}}\right]_{3}^{9}$

\begin{aligned} &=-\cos \frac{\pi}{2}+\cos 0+3-\frac{\pi}{2}+\frac{e^{9}-e^{3}}{e^{3}} \\ & \end{aligned}

$=0+1+3-\frac{\pi}{2}+\frac{e^{3}\left(e^{6}-1\right)}{e^{3}} \\$

$=4-\frac{\pi}{2}+e^{6}-1 \\$

$=3-\frac{\pi}{2}+e^{6}$