Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_4^9 \frac{\sqrt{x}}{\left( 30 - x^\frac{3}{2} \right)^2} d x . Then, \]
\[Let \left( 30 - x^\frac{3}{2} \right) = t . Then, - \frac{3}{2}\sqrt{x} dx = dt\]
\[When\, x = 4, t = 22\ and\ x\ = 9, t = 3\]
\[ \therefore I = \int_{22}^3 - \frac{2}{3}\frac{1}{t^2} dt\]
\[ \Rightarrow I = \frac{2}{3} \left[ \frac{1}{t} \right]_{22}^3 \]
\[ \Rightarrow I = \frac{2}{3}\left( \frac{1}{3} - \frac{1}{22} \right)\]
\[ \Rightarrow I = \frac{19}{99}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
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
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
