Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^1 \frac{1 - x}{1 + x} d\ x\ . Then, \]
\[I = \int_0^1 \left( \frac{1}{1 + x} - \frac{1 + x - 1}{1 + x} \right) d x\]
\[I = \int_0^1 \left( \frac{1}{1 + x} - 1 + \frac{x}{1 + x} \right) d x\]
\[ \Rightarrow I = \left[ \log \left( 1 + x \right) - x + \log \left( 1 + x \right) \right]_0^1 \]
\[ \Rightarrow I = \left( \log 2 - 1 + \log 2 \right) - \left( \log 1 - 0 + \log 1 \right)\]
\[ = 2 \log 2 - 1\]
APPEARS IN
RELATED QUESTIONS
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Choose the correct alternative:
Γ(1) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
