# How do you evaluate the integral of #int (3 - 2x)dx# from -1 to 3?

Here is a geometric approach.

Consider the graph of

The integral is the area under the curve above the

In the image below, the integral is the blue area minus the red area.

The blue triangle has base

The red triangle has base

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To evaluate the integral of a simple polynomial, simply do the inverse of the power rule for differentiation by adding one to the exponent, and then dividing by that new exponent. Then, find the value of the function by subtracting the function at the smaller bound from the function evaluated at the larger bound.

For your example, integrating we get:

Now substitute the bounds in:

Simplifying we get:

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If you do not yet have the Fundamental Theorem of Calculus available to evaluate this, but you need to evaluate it from a definition, see below.

I prefer to do this type of problem one small step at a time.

(We used summation formulas for the sums in the previous step.)

So,

There are several ways to think about this limit :

OR

Completing the integration

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To evaluate the integral ∫(3 - 2x)dx from -1 to 3, you first need to integrate the expression (3 - 2x) with respect to x and then evaluate it at the upper limit (3) and subtract the value obtained at the lower limit (-1).

∫(3 - 2x)dx = [3x - x^2] evaluated from -1 to 3

Substituting the upper limit (3) into the expression gives: 3(3) - (3)^2 = 9 - 9 = 0

Substituting the lower limit (-1) into the expression gives: 3(-1) - (-1)^2 = -3 - 1 = -4

Therefore, the value of the integral from -1 to 3 is 0 - (-4) = 4.

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