How do two different genres address a similar topic?

Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis

Tasya [4]

Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5

A.
$\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx$

B.
sepearte the integrals
$\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}$

next one
$\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}$

the last one you can do yourself, it is $\frac{50}{3}$
the sum is $\frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}$

so the area under the curve is $\frac{233}{3}$

6 0

2 years ago

A portion of a hiking trail slopes downward at about an 84° angle. To the

tangare [24]

C because if you divide the angle minus the feet you get 120

8 0

3 years ago

Read 2 more answers

Please answer correctly !!!!!! Will mark brainliest answer !!!!!!!!!!!!!!

AlekseyPX

Answer:

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