Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\, I = \int_{- 2}^1 \frac{\left| x \right|}{x} d x\]
\[\text{We have}, \]
\[\left| x \right| = \begin{cases}x&,& 0 \leq x \leq 1\\ - x&,& - 2 \leq x < 0\end{cases}\]
\[ \therefore \frac{\left| x \right|}{x} = \begin{cases}1&,& 0 \leq x \leq 1\\ - 1&,& - 2 \leq x < 0\end{cases}\]
\[\text{Therefore}, \]
\[I = \int_{- 2}^0 - 1dx + \int_0^1 1 dx\]
\[ = - \left[ x \right]_{- 2}^0 + \left[ x \right]_0^1 \]
\[ = 0 - 2 + 1 - 0\]
\[ = - 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate each of the following integral:
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
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_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
Γ(1) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
