Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 \frac{x}{x + 1} d x . Then, \]
\[I = \int_0^1 1 - \frac{1}{x + 1} d x\]
\[ \Rightarrow I = \left[ x - \log \left( x + 1 \right) \right]_0^1 \]
\[ \Rightarrow I = 1 - \log 2 - (0 - \log 1)\]
\[ \Rightarrow I = \log e - \log 2\]
\[ \Rightarrow I = \log \frac{e}{2}\]
APPEARS IN
संबंधित प्रश्न
Solve each of the following integral:
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
