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
Evaluate each of the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
Solve each of the following integral:
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Find: `int logx/(1 + log x)^2 dx`
