Definite Integration Question 40
Question 40
- Statement I The value of the integral
(2013 Main)
$$ \int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}} \text { is equal to } \pi / 6 \text {. } $$
Statement II $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
(a) Statement I is correct; Statement II is correct; Statement II is a correct explanation for Statement I
(b) Statement I is correct; Statement II is correct; Statement II is not a correct explanation for Statement I
(c) Statement I is correct; Statement II is false
(d) Statement I is incorrect; Statement II is correct
Passage Based Questions
Passage I
Let $F: R \rightarrow R$ be a thrice differentiable function. Suppose that $F(1)=0, F(3)=-4$ and $F^{\prime}(x)<0$ for all $x \in(1,3)$. Let $f(x)=x F(x)$ for all $x \in R$.
(2015 Adv.)
Show Answer
Solution:
- Let $I=\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}$
$$ \begin{array}{rlrl} \therefore & I & =\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan \frac{\pi}{2}-x}} \ & =\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}} \ \Rightarrow \quad & I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\tan x} d x}{1+\sqrt{\tan x}} \end{array} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{array}{rlrl} & & 2 I & =\int_{\pi / 6}^{\pi / 3} d x \ \Rightarrow & 2 I & =[x]_{\Pi / 6}^{\pi / 3} d x \ \Rightarrow & I & =\frac{1}{2} \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{12} \end{array} $$
Statement I is false.
But $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$ is a true statement by property of definite integrals.
