Advertisements
Advertisements
प्रश्न
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
Advertisements
उत्तर
\[\int_1^2 \frac{1}{x^2} e^\frac{- 1}{x} d x\]
\[Let \frac{- 1}{x} = t, then \frac{1}{x^2} dx = dt\]
\[\text{When, }x \to 1 ; t \to - 1\]
\[\text{And }x \to 2 ; t \to \frac{- 1}{2}\]
Therefore the integral becomes
\[ \int_{- 1}^\frac{- 1}{2} e^t d t\]
\[ = \left[ e^t \right]_{- 1}^\frac{- 1}{2} \]
\[ = e^\frac{- 1}{2} - e^{- 1} \]
\[ = \frac{\sqrt{e} - 1}{e}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
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 : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
If n > 0, then Γ(n) 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 ______.
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
`int x^3/(x + 1)` is equal to ______.
What does a definite integral represent?
