Advertisements
Advertisements
Question
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
Solution
\[I = \int_{- a}^a f\left( x^2 \right) d x\]
\[Here\ g\left( x \right) = f( x^2 )\]
\[ \Rightarrow g\left( - x \right) = f \left( - x \right)^2 = f( x^2 ) = g\left( x \right) i.e, g\left( x \right) \text{is even} \]
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
RELATED QUESTIONS
If f(2a − x) = −f(x), prove that
Evaluate each of the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
