By using the properties of definite integrals | vedclass

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

Question

(A) Let $I = \int_{0}^{2} x \sqrt{2-x} \, dx$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get: $I = \int_{0}^{2} (2-x

Vedclass · Accepted Answer

$16\sqrt{2}/15$

Vedclass · Answer

(A) Let $I = \int_{0}^{2} x \sqrt{2-x} \, dx$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get: $I = \int_{0}^{2} (2-x) \sqrt{2-(2-x)} \, dx = \int_{0}^{2} (2-x) \sqrt{x} \, dx$. $I = \int_{0}^{2} (2x^{1/2} - x^{3/2}) \, dx$. Integrating term by term: $I = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} \right]_{0}^{2} = \left[ \frac{4}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right]_{0}^{2}$. Substituting the limits: $I = \left( \frac{4}{3} (2)^{3/2} - \frac{2}{5} (2)^{5/2} \right) - 0$. Since $(2)^{3/2} = 2\sqrt{2}$ and $(2)^{5/2} = 4\sqrt{2}$: $I = \frac{4}{3} (2\sqrt{2}) - \frac{2}{5} (4\sqrt{2}) = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5}$. Taking the common denominator: $I = \frac{40\sqrt{2} - 24\sqrt{2}}{15} = \frac{16\sqrt{2}}{15}$.

By using the properties of definite integrals,evaluate the integral $\int_{0}^{2} x \sqrt{2-x} \, dx$.

Similar Questions

For $x > 0,$ if $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt,$ then $f(e) + f\left(\frac{1}{e}\right)$ is equal to:

If $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$,then the value of $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ is

The value of the integral $\int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$ is equal to:

If $I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \, d\theta$,then $I_{12} + I_{10} =$

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x + 1) \, dx$ is . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module