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
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f is an integrable function, show that
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
Find: `int logx/(1 + log x)^2 dx`
