Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{We have}, \]
\[I = \int\limits_1^2 \log_e \left[ x \right] dx\]
\[\text{We know that}, \]
\[\left[ x \right] = 1\text{, when }1 < x < 2\]
\[ \therefore I = \int\limits_1^2 \log_e 1 dx\]
\[I = \int\limits_1^2 \left( 0 \right) dx\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Evaluate the following:
Γ(4)
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
`int x^3/(x + 1)` is equal to ______.
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
