Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
Advertisements
उत्तर
We have,
\[\left| 2x - 1 \right| = \begin{cases} - \left( 2x - 1 \right)&,& 0 \leq x \leq \frac{1}{2}\\ 2x - 1&,& \frac{1}{2} \leq x \leq 1\end{cases}\]
\[ \therefore \int_0^1 \left| 2x - 1 \right| d x\]
\[ = \int_0^\frac{1}{2} - \left( 2x - 1 \right) dx + \int_\frac{1}{2}^1 \left( 2x - 1 \right) dx\]
\[ = \left[ - x^2 + x \right]_0^\frac{1}{2} + \left[ x^2 - x \right]_\frac{1}{2}^1 \]
\[ = \frac{- 1}{4} + \frac{1}{2} + 1 - 1 - \frac{1}{4} + \frac{1}{2}\]
\[ = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
Write the coefficient a, b, c of which the value of the integral
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "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
