Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{We have}, \]
\[I = \int\limits_0^{1 . 5} \left[ x \right] dx\]
\[ = \int_0^1 \left[ x \right] dx + \int_1^{1 . 5} \left[ x \right] dx\]
\[ = \int_0^1 \left( 0 \right) dx + \int_1^{1 . 5} \left( 1 \right)dx ................\left[\because \left[ x \right] = \begin{cases}0&& 0 \leq x < 1\\1&& 1 \leq x < 1 . 5\end{cases} \right]\]
\[ = 0 + \left[ x \right]_1^{1 . 5} \]
\[ = 1 . 5 - 1\]
\[ = 0 . 5\]
\[ = \frac{1}{2}\]
APPEARS IN
RELATED QUESTIONS
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
