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

$\int_0^{2a} \frac{f(x)}{f(x) + f(2a - x)} \, dx = $

Let $[\bullet]$ be the greatest integer function. If $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx$,then $\frac{1}{\pi} \int_{0}^{\alpha\pi} \left( \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} \right) d\theta$ is equal to . . . . . . .

The integral value $\int_{-2}^{0} [x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)] \, dx$ is

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$.

Value of the definite integral $\int_{-3}^{1} (2(t+1)^5 - 5(t+1)^3 + t + 3) dt$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module