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, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Prove that:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "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"d"x)/(x^4 + x^2 - 2)`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
