Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_4^9 \frac{1}{\sqrt{x}} d x . Then, \]
\[I = 2 \int_4^9 \frac{1}{2\sqrt{x}} d x\]
\[ \Rightarrow I = 2 \left[ \sqrt{x} \right]_4^9 \]
\[ \Rightarrow I = 2\left( 3 - 2 \right)\]
\[ \Rightarrow I = 2\]
APPEARS IN
संबंधित प्रश्न
If f(x) is a continuous function defined on [−a, a], then prove that
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
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 \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
