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Sketch the region enclosed by y=10x and y=x^2+24. Then find the area of the region.

How to get this answer?

User Npellow
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1 Answer

4 votes

Answer:

The area enclosed by given curves is
(4)/(3) square units.

Explanation:

The given equation are


y=10x


y=x^2+24

Draw the graph of both equation.

From the below graph it is clear that the graph intersect each other at (4,40) and (6,60).

The area of region enclosed by curves is


A=\int_(4)^(6)10xdx-\int_(4)^(6)(x^2+24)dx


A=10((1)/(2)[x^2]_(4)^(6))-(1)/(3)[x^3]_(4)^(6)-24[x]_(4)^(6)


A=5\left(6^2-4^2\right)-(1)/(3)\left(6^3-4^3\right)-24\left(6-4\right)


A=5[20]-(1)/(3)[152]-24[2]


A=100-(152)/(3)-48


A=(4)/(3)

Therefore the area enclosed by given curves is
(4)/(3) square units.

Sketch the region enclosed by y=10x and y=x^2+24. Then find the area of the region-example-1
User Zlatomir
by
6.8k points