Advertisements
Advertisements
प्रश्न
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
Advertisements
उत्तर
The given definite integral = `int_(-1)^2|x(x - 1)(x - 2)|dx`
= `int_(-1)^0 |x(x - 1)(x - 2)|dx + int_0^1 |x(x - 1)(x - 2)|dx + int_1^2 |x(x - 1)(x - 2)|dx`
= `- int_(-1)^0 (x^3 - 3x^2 + 2x)dx + int_0^1 (x^3 - 3x^2 + 2x)dx - int_1^2 (x^3 - 3x^2 + 2x)dx`
= `- [x^4/4 - x^3 + x^2]_(-1)^0 + [x^4/4 - x^3 + x^2]_0^1 - [x^4/4 - x^3 + x^2]_1^2`
= `9/4 + 1/4 + 1/4 = 11/4`
APPEARS IN
संबंधित प्रश्न
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
