Definite Integration Question 17
Question 17 - 31 January - Shift 1
Let a differentiable function $f$ satisfy $f(x)+\int_3^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$. Then $12 f(8)$ is equal to:
(1) 34
(2) 19
(3) 17
(4) 1
Show Answer
Answer: (3)
Solution:
Formula: Leibnitz Theorem, Integration by substitution, Standard formulas for Indefinite Integration
Differentiate w.r.t. $x$
$f^{\prime}(x)+\frac{f(x)}{x}=\frac{1}{2 \sqrt{x+1}}$
I.F. $=e^{\int \frac{1}{x} dx}=e^{\ln x}=x$
$x f(x)=\int \frac{x}{2 \sqrt{x+1}} d x$
Substitute $x+1=t^{2}$
$x f(x)=\int \frac{t^{2}-1}{2 t} 2 t d t$
$x f(x)=\frac{t^{3}}{3}-t+c$
$x f(x)=\frac{(x+1)^{3 / 2}}{3}-\sqrt{x+1}+c$
Also putting $x=3$ in given equation $f(3)+0=\sqrt{4}$
$f(3)=2$
$\Rightarrow C=8-\frac{8}{3}=\frac{16}{3}$
$f(x)=\frac{\frac{(x+1)^{3 / 2}}{3}-\sqrt{x+1}+\frac{16}{3}}{x}$
$f(8)=\frac{9-3+\frac{16}{3}}{8}=\frac{34}{24}$
$\Rightarrow 12 f(8)=17$
