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vovikov84 [41]
3 years ago
15

Write a definite integral that represents the area of the region. (Do not evaluate the integral.)

Mathematics
1 answer:
joja [24]3 years ago
8 0

Answer:

\int_{-1}^{1}7(x^{3}-x)\:dx

Step-by-step explanation:

1)  The other curve is y=0 then the common points of both curves are x-intercepts, the roots of y=7(x^{3}-x)

y=7(x^{3}-x)\Rightarrow 7(x^3-x)=0 \Rightarrow 7(x^{3}-x)=7x(x-1)(x-1)\Rightarrow \\S=\left ( 0,0 \right ),\left ( -1,0 \right ),\left ( 1,0 \right )

2). Then those intersection points are the upper and the lower limits. Plugging in to this formula for they belong to the interval [-1,1]:

\int_{a}^{b}|f(x)-g(x)dx

\int_{a}^{b}|f(x)-g(x)dx \Rightarrow \int_{-1}^{1}|7(x^{3}-x)-0|dx \Rightarrow \int_{-1}^{1}7(x^{3}-x)\:dx

 

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