Advertisements
Advertisements
प्रश्न
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
Advertisements
उत्तर
\[\int_1^3 \left| x^2 - 4 \right| d x\]
\[ = \int_1^2 - \left( x^2 - 4 \right) dx + \int_2^3 \left( x^2 - 4 \right) dx\]
\[ = \left[ - \frac{x^3}{3} + 4x \right]_1^2 + \left[ \frac{x^3}{3} - 4x \right]_2^3 \]
\[ = \frac{- 8}{3} + 8 + \frac{1}{3} - 4 + 9 - 12 - \frac{8}{3} + 8\]
\[ = 4\]
APPEARS IN
संबंधित प्रश्न
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
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/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following:
Γ(4)
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
`int x^3/(x + 1)` is equal to ______.
Which notation represents a definite integral?
Which integral has lower and upper limits?
Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?
