Question

Write a definite integral that represents the area of the region. (Do not evaluate the integral.)
y1 = 7(x3 − x)
y2 = 0

Answer 1

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]:

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

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