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
Evaluate the following integral:
Evaluate each of the following integral:
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Which integral has Constant C present?
What is an antiderivative of \[x\]?
Evaluate \[\int_{0}^{1}x\,dx\].
What are definite integrals used to find over a fixed interval?
