Advertisements
Advertisements
प्रश्न
If f(2a − x) = −f(x), prove that
Advertisements
उत्तर
Using additive property
\[I = \int_0^a f\left( x \right) d x + \int_a^{2a} f\left( x \right) d x\]
\[\text{Consider the integral} \int_a^{2a} f\left( x \right) d x\]
\[\text{Let }x = 2a - t, \text{Then }dx = - dt\]
\[\text{When }x = a, t = a\text{ and }x = 2a, t = 0\]
Therefore,
\[ \int_a^{2a} f\left( x \right) d x = - \int_a^0 f\left( 2a - t \right) d t\]
\[ = \int_0^a f\left( 2a - t \right) d t\]
\[ = \int_0^a f\left( 2a - x \right) dx ................\left( \text{changing the variable} \right)\]
\[\text{We have }f\left( 2a - x \right) = - f\left( x \right)\]
Therefore,
\[I = \int_0^a f\left( x \right) d x - \int_0^a f\left( x \right) d x = 0\]
APPEARS IN
संबंधित प्रश्न
If f is an integrable function, show that
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
