Advertisements
Advertisements
Question
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
Advertisements
Solution
\[\int_1^2 x\sqrt{3x - 2} d x\]
\[Let, 3x - 2 = t,\text{ then }3dx = dt\]
\[\text{when, }x = 1 ; t = 1\text{ and }x = 2 ; t = 4\]
\[\text{Therefore the integral becomes}\]
\[ \int_1^4 \frac{t + 2}{3}\sqrt{t} \frac{dt}{3}\]
\[ = \frac{1}{9} \int_1^4 t^\frac{3}{2} + 2\sqrt{t} dt\]
\[ = \frac{1}{9} \left[ \frac{2 t^\frac{5}{2}}{5} + \frac{4 t^\frac{3}{2}}{3} \right]_1^4 \]
\[ = \frac{1}{9}\left[ \frac{64}{5} + \frac{32}{3} - \frac{2}{5} - \frac{4}{3} \right]\]
\[ = \frac{46}{135}\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
