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Svetlanka [38]
2 years ago
12

Evaluate the integral (3 - 2x)dx for x = -1 to x = 3 byinterpreting it in terms of areas.

Mathematics
1 answer:
suter [353]2 years ago
5 0

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.

\int^{3}_{-1} {3-2x} \, dx \\\left|3x - 2\dfrac{x^2}{2}\right|^{3}_{-1}\\\left|3x - x^2\right|^{3}_{-1}

we have integrated the equation, and now we're going to put the limits find the area under the function f(x)

\left|3x - x^2\right|^{3}_{-1}\\(3(3)-(3)^2)-(3(-1)-(-1)^2)

-------------

if these seems like a big jump, try to understand it through this:

\left|3x - x^2\right|^{b}_a\\(3(b)-(b)^2)-(3(a)-(a)^2)\\

we've only distributed the limits each time to the same integrated expression.

--------------

coming back to our solution:

(3(3)-(3)^2)-(3(-1)-(-1)^2)

(0)-(-4)

4

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. <u>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</u><u>. In other words, the net area is above the x-axis and that is equal to 4 unit^2 </u>

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