Advertisements
Advertisements
प्रश्न
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
पर्याय
- \[\frac{\pi}{2}\]
- \[\frac{1}{2}\]
- \[\frac{\pi}{4}\]
1
Advertisements
उत्तर
\[\int_0^\alpha \frac{1}{1 + 4 x^2} d x = \frac{\pi}{8}\]
\[ \Rightarrow \int_0^\alpha \frac{1}{1 + \left( 2x \right)^2} d x = \frac{\pi}{8}\]
\[ \Rightarrow \frac{1}{2} \left[ \tan^{- 1} 2x \right]_0^\alpha = \frac{\pi}{8}\]
\[ \Rightarrow \frac{1}{2} \tan^{- 1} 2\alpha = \frac{\pi}{8}\]
\[ \Rightarrow 2\alpha = \tan\frac{\pi}{4}\]
\[ \Rightarrow 2\alpha = 1\]
\[ \therefore \alpha = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "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:
Γ(1) is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
