#### To determine

**To:**

(i) Evaluate the integral,

(ii) Interpret the integral as a difference of areas.

#### Answer

∫-12x3 dx=154

#### Explanation

**1) Concept:**

i) Fundamental theorem of Calculus, Part 2

If f is continuous on a,b, then

∫abfxdx=Fb-F(a)

where F is any antiderivative of f, that is, a function F such that F'=f

ii) Power rule for antiderivative:

ddxxn+1n+1=xn

**2) Given:**

y=x3, -1≤x≤2

**3) Calculation:**

(i)

The function fx=x3 is continuous on -1,2

By using concept i) (Fundamental theorem of Calculus, Part 2),

∫-12x3 dx=F2-F-1…(1)

Where F is antiderivative of f, that is, a function F such that F'(x)=f(x) means

ddxFx=fx…(2)

By using concept ii) (power rule of antiderivative),

ddxx3+13+1=x3

ddxx44=x3

From (2),

Fx=x44

Substitute F(x) in (1) at x=-1 and x=2,

∫-12x3 dx=F2-F(-1)

=244--144

=164-14

=16-14

=154

Therefore,

∫-12x3 dx=154

(ii) To interpret integral as a difference of areas

The region enclosed by the curve y=x3, x=-1, x=2 and y=0 is shown by the black region

The region enclosed by the curve y=x3, x=-1, x=2 and y=0 is shown by the black region is the difference of area above x-axis and area below x-axis

That is, the difference of area between x=0 to x=2 and area between x=-1 to 0

**Conclusion:**

Therefore, the required area is A=154