Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
Advertisements
उत्तर
\[\text{We have}, \]
\[I = \int\limits_0^1 \left\{ x \right\} dx\]
\[\text{We know} \left\{ x \right\} = x, 0 < x < 1\]
\[ \therefore I = \int\limits_0^1 x\ dx\]
\[ = \left[ \frac{x^2}{2} \right]_0^1 \]
\[ = \frac{1}{2} - \frac{0}{2}\]
\[ = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
Evaluate each of the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_2^3 e^{- x} dx\]
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_0^1 log (1/x - 1) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
