Respuesta :
Answer:
4 unit^2
explanation is given at the end.
Step-by-step explanation:
What this integral represents is the net area between the function f(x) = 3 - 2x
and the x-axis, between the range of x between -1 and 3.
[tex]\int^{3}_{-1} {3-2x} \, dx \\\left|3x - 2\dfrac{x^2}{2}\right|^{3}_{-1}\\\left|3x - x^2\right|^{3}_{-1}[/tex]
we have integrated the equation, and now we're going to put the limits find the area under the function f(x)
[tex]\left|3x - x^2\right|^{3}_{-1}\\(3(3)-(3)^2)-(3(-1)-(-1)^2)[/tex]
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if these seems like a big jump, try to understand it through this:
[tex]\left|3x - x^2\right|^{b}_a\\(3(b)-(b)^2)-(3(a)-(a)^2)\\[/tex]
we've only distributed the limits each time to the same integrated expression.
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coming back to our solution:
[tex](3(3)-(3)^2)-(3(-1)-(-1)^2)[/tex]
[tex](0)-(-4)[/tex]
[tex]4[/tex]
so our area is 4
the area above the x-axis is positive, and the area below the x-axis is negative. Since, our answer is +4. We now know that if within this range [-1,3] the area above and below the x-axis exist, there is more area above the x-axis than below the x-axis. In other words, the net area is above the x-axis and that is equal to 4 unit^2
