Advertisements
Advertisements
Question
`int_0^(2a)f(x)dx`
Advertisements
Solution
We have.
`I=int_0^(2a)f(x)dx`
Then,
`I=int_0^af(x)dx+int_a^(2a)f(x)dx`
`I=int_0^af(x)dx+I_1 ......................["where, "I_1=int_a^(2a)f(x)dx]`
Let 2a - t = x then dx = - dt
If t = a ⇒ x = a
If t = 2a ⇒ x = 0
`I_1=int_0^(2a)f(x)dx=int_a^0f(2a-t)(-dt)=-int_a^0f(2a-t)dt`
`I_1=int_0^af(2a-t)dt=int_0^af(2a-x)dx`
`thereforeI=int_0^af(x)dx+int_0^af(2a-x)dx`
`I=int_0^af(x)dx+int_0^af(x)dx=2int_0^af(x)dx............................[f(2a-x=f(x))]`
Hence proved.
APPEARS IN
RELATED QUESTIONS
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
