How do you find the area cut off by the x axis, above the x-axis, and #y=(3+x)(4-x)#?

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Answer 1 · 2017-01-13T21:02:23

#343/6# square units.

Start by expanding.

#y = -x^2 + 4x - 3x + 12#

#y = -x^2 + x + 12#

This is a parabola that opens downward. So, the area we need to find is between #a# and #b#, where #a# and #b# are the zeroes of your function.

We started in factored form, so we can immediately say that our zeroes are #x = -3# and #x= 4#.

We need to evaluate the following integral:

#int_-3^4 -x^2 + x + 12 dx#

#=int_-3^4 -1/3x^3 + 1/2x^2 + 12x #

Use the second fundamental theorem of calculus to evaluate.

#=-1/3(4)^3 + 1/2(4)^2 + 12(4) - (-1/3(-3)^3 + 1/2(-3)^2 + 12(-3))#

#= -64/3 + 8 + 48 - 9 - 9/2 +36#

#=11 - 9/2 - 64/3#

#=83 - 9/2 - 64/3#

#=343/6# square units

Hopefully this helps!