Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_{- 1}^1 5 x^4 \sqrt{x^5 + 1}\ d\ x . Then, \]
\[Let\ x^5 + 1 = t . Then, 5 x^4\ dx = dt\]
\[When\ x = - 1, t = 0\ and\ x = 1, t = 2\]
\[ \therefore I = \int_0^2 \sqrt{t}\ dt\]
\[ \Rightarrow I = \left[ \frac{2}{3} t^\frac{3}{2} \right]_0^2 \]
\[ \Rightarrow I = \frac{2}{3}\sqrt{8}\]
\[ \Rightarrow I = \frac{4\sqrt{2}}{3}\]
APPEARS IN
संबंधित प्रश्न
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
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
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
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
Γ(4)
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
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: