Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^a \sqrt{a^2 - x^2} d x . \]
\[Let\ x = a\ \sin\ t . Then\, dx = a\ \cos\ t\ dt\]
\[When\ x = 0, t = 0\ and\ x\ = a, t = \frac{\pi}{2}\]
\[ \therefore I = \int_0^a \sqrt{a^2 - x^2} d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \sqrt{\left( a^2 - a^2 \sin^2 t \right)} a \cos\ t\ d\ t\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} a^2 \cos^2 t\ dt\]
\[ \Rightarrow I = a^2 \int_0^\frac{\pi}{2} \frac{1 + \cos 2t}{2} dt\]
\[ \Rightarrow I = \frac{a^2}{2} \left[ t + \frac{\sin 2t}{2} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{a^2}{2}\left( \frac{\pi}{2} - 0 \right)\]
\[ \Rightarrow I = \frac{\pi^2}{4}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate each of the following integral:
If f is an integrable function, show that
Prove that:
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
