Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{We have}, \]
\[I = \int\limits_0^1 2^{x - \left[ x \right]} dx\]
\[ = \int\limits_0^1 2^{x - 0} dx ...............\left( \because \left[ x \right] = 0\text{ where, }0 < x < 1 \right)\]
\[ = \int\limits_0^1 2^x dx\]
\[ = \left[ \frac{2^x}{\log_e 2} \right]_0^1 \]
\[ = \frac{2^1}{\log_e 2} - \frac{2^0}{\log_e 2}\]
\[ = \frac{2}{\log_e 2} - \frac{1}{\log_e 2}\]
\[ = \frac{1}{\log_e 2}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate each of the following integral:
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
Γ(n) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
