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
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Evaluate each of the following integral:
Evaluate each of the following integral:
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
