3 years ago

A technician charges $25 per hour plus $50 for a

Mathematics

1 answer:

Usimov [2.4K]3 years ago

8 0

Hmmmmmmmmmmmmmmmmmmmmm

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(10 * 7) / 2 =
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Dafna11 [192]

Anna= a
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3 years ago

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

joja [24]

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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Step-by-step explanation:

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