Find the area of the region (in square units) enclosed by the curves: y2=8(x+2),y2=4(1βx) y^2 = 8(x+2), \quad y^2 = 4(1-x) y2=8(x+2),y2=4(1βx) and the Y-axis.
Evaluate the integral: I=β«β33β£2βxβ£dx. I = \int_{-3}^{3} |2 - x| dx. I=β«β33ββ£2βxβ£dx.
Evaluate the integral: I=β«βΟΟxsinβ‘3x4βcosβ‘2xdx. I = \int_{-\pi}^{\pi} \frac{x \sin^3 x}{4 - \cos^2 x} dx. I=β«βΟΟβ4βcos2xxsin3xβdx.
If β«32cosβ‘3x2sinβ‘2xdx=32(tanβ‘x)Ξ²+310(tanβ‘x)4+C \int \frac{3}{2\cos 3x \sqrt{2} \sin 2x} dx = \frac{3}{2} (\tan x)^{\beta} + \frac{3}{10} (\tan x)^4 + C β«2cos3x2βsin2x3βdx=23β(tanx)Ξ²+103β(tanx)4+C then A= A = A= ?