int_{1}^{3} frac{sqrt{4-x}}{sqrt{x}+sqrt{4-x} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $I = \int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx \quad \dots(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = 1+3 = 4$.Substituting $x$ with $(4-x)$:$I = \int_{1}^{3} \frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}} dx$$I = \int_{1}^{3} \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} dx \quad \dots(ii)$Adding equations $(i)$ and $(ii)$:$2I = \int_{1}^{3} \left( \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} + \frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}} \right) dx$$2I = \int_{1}^{3} \frac{\sqrt{4-x}+\sqrt{x}}{\sqrt{x}+\sqrt{4-x}} dx$$2I = \int_{1}^{3} 1 dx$$2I = [x]_{1}^{3} = 3 - 1 = 2$$I = \frac{2}{2} = 1$

$\int_{1}^{3} \frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}} dx$ is equal to

Similar Questions

If $f$ and $g$ are continuous functions in $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 4$,then $\int_{0}^{a} f(x) g(x) dx$ is equal to:

If $f(t) = \int_0^t \tan^{(2n-1)} x \, dx$,$n \in N$,then $f(t+\pi) =$

$\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x=$

The value of the integral $\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$,where $p$ and $q$ are integers,is equal to:

If $(a, b)$ is the orthocentre of the triangle whose vertices are $(1, 2), (2, 3)$ and $(3, 1)$,and $I_1 = \int_{a}^{b} x \sin(4x - x^2) dx$,$I_2 = \int_{a}^{b} \sin(4x - x^2) dx$,then $36 \frac{I_1}{I_2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module