# `int_-1^2(x^3)dx` Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch.

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`int_-1 ^2(x^3)dx `

Let us first solve:

`int (x^3)dx = ((x^(3+1)) /(3+1) )+c = (x^4 /4 )+C`

Looking at the graph below we can say that the area covered by the curve `x^3` for `-1<=x<=2 ` is equal to `int_-1 ^2(x^3)dx`

= area covered below x -axis with `-1<=x<=0` + area covered above x -axis with `0<=x<=2`

= `int_-1^0 x^3 dx `+ int _0^2 x^3 dx`

as we know `int x^3 dx = x^4/4`

so,

=`[x^4/4]_-1 ^0 +[x^4]_0 ^2`

=`[0]-[(-1)^4/4] +[2^4/4]-[0]`

= `-1/4 +4`

= `15/4 `

Or we can solve it using limits as follows:

`int_ a ^b f(x)dx = lim_(x->b-) f'(x) -lim_(x->a+) f'(x)`

so now let us evaluate as follows,

`int_-1 ^2(x^3)dx` where f'(x) =x^4/4 a= -1 ,b =2`

so ,

`int_-1 ^2(x^3)dx= lim_(x->(2)-) (x^4)/4 -lim_(x->(-1)+) (x^4)/4`

let us solve individually the above,

`lim_(x->(2)-) (x^4)/4 `

=> let us put the value `x= 2`

so ,we get

`lim_(x->(2)-) (2^4)/4 = 4`

and

`lim_(x->(-1)+) (x^4)/4`

put the value x=-1,so we get

`lim_(x->(-1)+) (-1^4)/4 = 1/4` ,so now

`int_-1 ^2(x^3)dx`

= `lim_(x->(2)-) (x^4)/4 -lim_(x->(-1)+) (x^4)/4 = 4-1/4 `

= `15/4`

Find the antiderivative of each term: `int (a^n=(a^(n+1))/(n+1))`

`int_-1^2(x^3)dx=x^4/4|_-1^2`

Plug in the upper value and subtract out the lower limit

`[(2)^4/4]-[(-1)^4/4]`

=3.75

The area above the x-axis subtracted from the area under the x-axis is 3.75