Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\, I = \int_{- 2}^1 \frac{\left| x \right|}{x} d x\]
\[\text{We have}, \]
\[\left| x \right| = \begin{cases}x&,& 0 \leq x \leq 1\\ - x&,& - 2 \leq x < 0\end{cases}\]
\[ \therefore \frac{\left| x \right|}{x} = \begin{cases}1&,& 0 \leq x \leq 1\\ - 1&,& - 2 \leq x < 0\end{cases}\]
\[\text{Therefore}, \]
\[I = \int_{- 2}^0 - 1dx + \int_0^1 1 dx\]
\[ = - \left[ x \right]_{- 2}^0 + \left[ x \right]_0^1 \]
\[ = 0 - 2 + 1 - 0\]
\[ = - 1\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
`int x^3/(x + 1)` is equal to ______.
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
