Advertisements
Advertisements
प्रश्न
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
Advertisements
उत्तर
\[\int_1^2 x\sqrt{3x - 2} d x\]
\[Let, 3x - 2 = t,\text{ then }3dx = dt\]
\[\text{when, }x = 1 ; t = 1\text{ and }x = 2 ; t = 4\]
\[\text{Therefore the integral becomes}\]
\[ \int_1^4 \frac{t + 2}{3}\sqrt{t} \frac{dt}{3}\]
\[ = \frac{1}{9} \int_1^4 t^\frac{3}{2} + 2\sqrt{t} dt\]
\[ = \frac{1}{9} \left[ \frac{2 t^\frac{5}{2}}{5} + \frac{4 t^\frac{3}{2}}{3} \right]_1^4 \]
\[ = \frac{1}{9}\left[ \frac{64}{5} + \frac{32}{3} - \frac{2}{5} - \frac{4}{3} \right]\]
\[ = \frac{46}{135}\]
APPEARS IN
संबंधित प्रश्न
Solve each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
Γ(n) is
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
