Advertisements
Advertisements
प्रश्न
Advertisements
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
If f is an integrable function, show that
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
