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Calculus- Early Transcendentals, 2021a

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306 Applications of Integration<br />

Guidelines for Area Between Two Curves<br />

1. Find the intersection points.<br />

2. Draw a sketch of the two curves.<br />

3. Using the sketch determine which curve is the top curve and which curve is the bottom curve.<br />

You may need to split the area up into multiple regions if the curves intersect multiple times<br />

in [a,b].<br />

4. Put the above information into the appropriate formula (once for each region):<br />

Area =<br />

∫ b<br />

a<br />

(top curve) − (bottom curve)dx, a ≤ x ≤ b.<br />

5. Evaluate the integral using the Fundamental Theorem of <strong>Calculus</strong> (you should get a positive<br />

number representing an area).<br />

Example 8.4: Area Between Two Curves<br />

Determine the area enclosed by y = x 2 , y = √ x, x = 0 and x = 2.<br />

Solution. The points of intersection of y = x 2 and y = √ x are<br />

x 2 = √ x → x 4 = x → x 4 − x = 0 → x(x 3 − 1)=0.<br />

Thus, either x = 0orx = 1. Sketching the curves gives:<br />

The area we want to compute is the shaded region. Since the top curve changes at x = 1, we need to<br />

use the formula twice. For A 1 we have a = 0, b = 1, the top curve is y = √ x and the bottom curve is y = x 2 .<br />

For A 2 we have a = 1, b = 2, the top curve is y = x 2 and the bottom curve is y = √ x.<br />

For the first integral we have:<br />

Area = A 1 + A 2 =<br />

∫ 1<br />

0<br />

∫ 1<br />

0<br />

( √ x − x 2 )dx =<br />

( √ x − x 2 )dx+<br />

∫ 2<br />

1<br />

( 2<br />

3 x3/2 − 1 3 x3 )∣ ∣∣∣<br />

1<br />

0<br />

(x 2 − √ x)dx<br />

= 1 3

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