1. Area Volume

1. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

2. Find the area of the region between and for .
area =????

3. Find the area of the region enclosed between and from to . Hint: Notice that this region consists of two parts.

2. Originally Posted by Kayla_N
1. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

2. Find the area of the region between and for .
area =????

3. Find the area of the region enclosed between and from to . Hint: Notice that this region consists of two parts.

1) If you graph the the two equations on a graphing device, you will notice that y = 5x is above y = 6x^2. Hence your integral will be with respect to x integrating (5x - 6x^2)dx between the two point of intersection. The intersection can be found by equation 6x^2 = 5x and solving for x.

2) Use the same method from the previous question but now your limits are from 0 to 1 with respect to x. Remember you are subtracting two areas essentially, so the bottom function which has a smaller area beneath it should be subtracted from the top with a larger area. This technique is easy to employ by again, using a graphing device.

3) Ill leave as an exercise to you hoping that the previous two have helped you

Please message if you need further clarification

3. Originally Posted by Kayla_N
1. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

2. Find the area of the region between and for .
area =????

3. Find the area of the region enclosed between and from to . Hint: Notice that this region consists of two parts.

$\displaystyle \int_0^{\frac{5}{6}} 5x \cdot dx - \int_0^{\frac{5}{6}} 6x^2 \cdot dx$
$\displaystyle \int_0^{\frac{5}{6}} 5x \cdot dx - \int_0^{\frac{5}{6}} 6x^2 \cdot dx$