Advertisements
Advertisements
प्रश्न
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
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
Evaluate the following integral:
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate :
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
Γ(4)
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
