Advertisements
Advertisements
प्रश्न
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Advertisements
उत्तर
\[I = \int_{- a}^a f\left( x^2 \right) d x\]
\[Here\ g\left( x \right) = f( x^2 )\]
\[ \Rightarrow g\left( - x \right) = f \left( - x \right)^2 = f( x^2 ) = g\left( x \right) i.e, g\left( x \right) \text{is even} \]
Therefore
\[I = 2 \int_0^a f\left( x^2 \right) d x .............\left[\text{Using }\int_{- a}^a g\left( x \right) d x = 2 \int_0^a g\left( x \right) dx \text{ when }g\left( x \right) \text{is even} \right]\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
