Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{Let I }=\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]
\[= \int_\frac{1}{3}^1 \frac{\left[ x^3 \left( \frac{x}{x^3} - 1 \right) \right]^\frac{1}{3}}{x^4}dx\]
\[ = \int_\frac{1}{3}^1 \frac{x \left( \frac{1}{x^2} - 1 \right)^\frac{1}{3}}{x^4}dx\]
\[ = \int_\frac{1}{3}^1 \frac{\left( \frac{1}{x^2} - 1 \right)^\frac{1}{3}}{x^3}dx\]
Put
\[\therefore - \frac{2}{x^3}dx = dz\]
\[ \Rightarrow \frac{dx}{x^3} = - \frac{dz}{2}\]
When
When
\[\therefore I = - \frac{1}{2} \int_8^0 z^\frac{1}{3} dz\]
\[ = \left.- \frac{1}{2} \times \frac{z^\frac{4}{3}}{\frac{4}{3}}\right|_8^0 \]
\[ = - \frac{3}{8}\left[ 0 - \left( 8 \right)^\frac{4}{3} \right]\]
\[ = - \frac{3}{8} \times \left( - 16 \right)\]
\[ = 6\]
APPEARS IN
RELATED QUESTIONS
Evaluate :
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
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
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following:
`Γ (9/2)`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
