I have another integral problem.?

Question

how do i find the integral between 3 and -2 (3 on top) for
x^3(x^2 - 1)dx

Answer 1

Multiply it out, then integrate normally:

Int[ x³(x² - 1), -2, 3]

Int[ x^5 - x³, -2, 3]

(1/6)x^6 - (1/4)x^4 |(-2,3)

(1/6)(3^6-(-2)^6) - (1/4)(3^4 - (-2)^4)

= 1135/12

Answer 2

First off, when you "read" an integral, you say from the bottom to the top: so you should say the integral from -2 to 3 of x^3(x^2-1)dx.

Now, for this integral, you should distribute the x^3, and then integrate this as two integrals:

=Int (-2, 3) [x^5 - x^3]
=1/6 x^6 - 1/4x^4 (-2, 3)
= 1/6 3^6 - 1/4 3^4 - (1/6 (-2)^6 - 1/6 (-2)^4)
= 1135/12

Answer 3

[-2, 3]∫x³(x²-1) dx
Distribute the x³:
[-2, 3]∫x^5-x^3 dx
Now integrate term by term:
x^6/6 - x^4/4 from -2 to 3
And evaluate:
729/6 - 81/4 - (64/6 - 16/4)
1458/12 - 243/12 - 128/12 + 48/12

1135/12

Answer 4

Multiply out the integrand to get x^5 - x^3, now you have the sum of two integrals: ∫x^5dx - ∫x^3dx, which result in (x^6)/6 - (x^4)/4 Evaluate at the limits to get (3^6)/6 - (3^4)/4 - (-2^6)/6 + (-2^4)/4; you can do the computations.

Answer 5

Expand the function before integrating.

∫x^3(x^2 - 1)dx =
∫(x^5 - x^3)dx =
. ......................... 3
(1/6)x^6 - (1/4)x^4 | =
............................ -2

729/6 - 81/4 - 64/6 + 16/4 =
665/6 - 65/4 =
(1330 - 195)/12 =
1135/12 =
94.5833333333

Answer 6

The 2d critical is high quality. to acquire ? [x/(x + 3)]dx placed u = x +3 Then dx = du and x = u - 3 critical will change into ? [(u - 3)/u]du = ? [ a million - 3/u]du = u -3 lnu = x + 3 - 3ln|x + 3| the completed critical is then x + 3 - 3ln|x + 3| - 3ln|x + 3| +C = x + 3 - 6ln|x + 3| +C. verify. If we differentiate the above, we arrive decrease back on the integrand d/dx[x + 3 - 6ln|x + 3| + C] = a million - 6/(x + 3) = (x + 3/(x + 3) - 6/(x + 3) = [x + 3 - 6] / (x + 3) = (x - 3) / (x + 3) that is the unique function.

Answer 7

Multiply it out into: x^5 - x^3.
Then you can do it in two different integrals:
int(x^5,-2,3) - int(x^3,-2,3)

Answer 8

=(1/6)X^6- (1/4)X^4 (3 on top)....(-2 on bottom)
=(1/6)(3)^6 - (1/4)(3)^4-(1/6)(-2)^6+(1/4)(-2)^4...............................................