Mathematics: Sketch the region enclosed by y=6x and y=x^2+8 then find the area of the region. how to get this ?

Sketch the region enclosed by y=6x and y=x^2+8 then

unknown6669

31.12.2021

The area enclosed by given curves is $\frac{4}{3}$ square units.

Step-by-step explanation:

The given equations are

y=6x

y=x^2+8

Draw the graph of given equations.

From the below graph it is clear that both curves intersect each other at (2,12) and (4,24).

The area of region enclosed by curves is

$A=\int_{2}^{4}6xdx-\int_{2}^{4}(x^2+8)dx$

$A=6(\frac{1}{2}[x^2]_{2}^{4})-\frac{1}{3}[x^3]_{2}^{4}-8[x]_{2}^{4}$

$A=3\left(4^2-2^2\right)-\frac{1}{3}\left(4^3-2^3\right)-8\left(4-2\right)$

$A=3[12]-\frac{1}{3}[56]-8[2]$

$A=36-\frac{56}{3}-16$

$A=\frac{4}{3}$

Therefore the area enclosed by given curves is $\frac{4}{3}$ square units.

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