Calculus definite integrals help again?

Question

Evaluate given definite integral

2X(4-X^2)^3 dx
upper = -1
lower = -2

Answer 1

I don't get the same answer as Mohit, so I'll show more details.

The antiderivative is (-1/4)*(4 - x^2)^4 . It is easily shown that the derivative of this is the original integrand. The evaluation from -2 to -1 is
(-1/4)[(4 - (-1)^2)^4 - (4 - (-2)^2)^4] = (-1/4)[(4 - 1)^4 - (4 - 4)^4]
= (-1/4)[3^4 - 0 ^4]
= (-1/4)*81
= -81/4 .

Answer 2

Ans:
-625/4
=-156.25



simple :
put 4-x^2 = t

Answer 3

f_avg=a million/(b-a) * INT(f(t)dt) from a to b, so f_avg=a million/(2-0) * INT((a million+t)dt) from 0 to 2 f_avg=a million/2 * (t+(t^2)/2) from 0 to 2 f_avg=2 yet another answer would be to attain that f(t)=a million+t is linear so f_avg=a million/(b-a)(f(b)+f(a)) f_avg=(a million/2)(3+a million) f_avg=2 area under curve from x=[a,b] is INT(f(x)dx) from a to b The arch of f(x)=sin(x) is x=[0,pi], so the area under the arch of f(x)=sin(x) is INT(sin(x)dx) from o to pi You get to try this one. bear in concepts, (d/dx)(cos(x)) = -sin(x)