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
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following integral:
Evaluate the following integral:
If f is an integrable function, show that
Write the coefficient a, b, c of which the value of the integral
Evaluate :
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^4 x dx\]
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Find: `int logx/(1 + log x)^2 dx`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
