Respuesta :

ALin03
 Sure! So if you want to find the area under a curve, this calls for an integral. I am assuming you know what that is, since this looks like a calculus question.

Since we cant to find the area under y=(x+1)^3 from -1 to 1, we can write the integral as "the integral of (x+1)^3 from -1 to 1", or seen as [tex] \int\limits^1_-1 {(x+1)^3} \, dx [/tex]
You can then plug this into your calculator and find it to be 4



The (maybe) geometry way: 
We can separate this graph into two parts: a triangle and a trapezoid using the dotted lines. Then, we can find the area of the triangle and trapezoid and add them together. The area of the triangle is 1/2(b)(h). The base is 1 and the height is -1, so therefore, the area of the triangle is -1/2.
The area of a trapezoid is 1/2(a+b)*h. The height is 1, the a (top part) is 1, and the b (bottom part) is 8. Therefore, we plug that in and get 4.5
4.5+(-0.5)=4
check the picture below.

[tex]\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\ --------\\ h=1\\ a=1\\ b=2 \end{cases}\\\\ -------------------------------\\\\ \stackrel{trapezoid's~area}{\cfrac{1(1+2)}{2}}~~~~+~~~~\stackrel{triangle's~area}{\cfrac{1}{2}(1)(7)}\implies \cfrac{3}{2}+\cfrac{7}{2}\implies \cfrac{3+7}{2}\implies 5[/tex]
Ver imagen jdoe0001
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