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

$\int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx = $

For $U_n = \int\limits_0^1 x^n (2 - x)^n \, dx$ and $V_n = \int\limits_0^1 x^n (1 - x)^n \, dx$,where $n \in N$,which of the following statements is true?

$\int_0^\pi (\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x) dx =$

$\int_0^\pi \sin^2 x \cos^3 x \, dx = $ . . . . . . .

If $\int_{0}^{2}(\sqrt{2x}-\sqrt{2x-x^{2}}) dx = \int_{0}^{1}(1-\sqrt{1-y^{2}}-\frac{y^{2}}{2}) dy + \int_{1}^{2}(2-\frac{y^{2}}{2}) dy + I$,then $I = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module