Advertisements
Advertisements
प्रश्न
Evaluate the following integral:
Advertisements
उत्तर
\[I = \int_{- 3}^3 \left| x + 1 \right| d x\]
\[We\ know\ that, \left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &, &- 3 \leq x \leq - 1 \\x + 1 &, &- 1 < x \leq 3\end{cases}\]
\[ \therefore I = \int_{- 3}^{- 1} - \left( x + 1 \right) d x + \int_{- 1}^3 \left[ x + 1 \right] d x\]
\[ \Rightarrow I = \left[ - \frac{\left( x + 1 \right)^2}{2} \right]_{- 3}^{- 1} + \left[ \frac{\left( x + 1 \right)^2}{2} \right]_{- 1}^3 \]
\[ \Rightarrow I = 0 + 2 + 8 - 0\]
\[ \Rightarrow I = 10\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
Evaluate the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
