Advertisements
Advertisements
Question
Evaluate the following integral:
Advertisements
Solution
\[I = \int_{- 3}^3 \left| x + 1 \right| d x\]
\[We\ know\ that, \left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &, &- 3 \leq x \leq - 1 \\x + 1 &, &- 1 < x \leq 3\end{cases}\]
\[ \therefore I = \int_{- 3}^{- 1} - \left( x + 1 \right) d x + \int_{- 1}^3 \left[ x + 1 \right] d x\]
\[ \Rightarrow I = \left[ - \frac{\left( x + 1 \right)^2}{2} \right]_{- 3}^{- 1} + \left[ \frac{\left( x + 1 \right)^2}{2} \right]_{- 1}^3 \]
\[ \Rightarrow I = 0 + 2 + 8 - 0\]
\[ \Rightarrow I = 10\]
APPEARS IN
RELATED QUESTIONS
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
If n > 0, then Γ(n) is
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
