Area of the region (in square units) bounded | vedclass

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(D) To find the area of the region bounded by the curve $y=x^2+4$ and the line $y=5x-2$, we first find their points of intersection by setting the equ

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$\frac{1}{6}$

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(D) To find the area of the region bounded by the curve $y=x^2+4$ and the line $y=5x-2$, we first find their points of intersection by setting the equations equal to each other:$x^2+4 = 5x-2$$x^2-5x+6 = 0$$(x-2)(x-3) = 0$Thus, the points of intersection are at $x=2$ and $x=3$.In the interval $[2, 3]$, the line $y=5x-2$ lies above the curve $y=x^2+4$.The area $A$ is given by the integral:$A = \int_2^3 ((5x-2) - (x^2+4)) \, dx$$A = \int_2^3 (5x - x^2 - 6) \, dx$$A = \left[ \frac{5x^2}{2} - \frac{x^3}{3} - 6x \right]_2^3$Evaluating at the limits:$A = \left( \frac{5(3)^2}{2} - \frac{(3)^3}{3} - 6(3) \right) - \left( \frac{5(2)^2}{2} - \frac{(2)^3}{3} - 6(2) \right)$$A = \left( \frac{45}{2} - 9 - 18 \right) - \left( 10 - \frac{8}{3} - 12 \right)$$A = \left( \frac{45}{2} - 27 \right) - \left( -2 - \frac{8}{3} \right)$$A = \left( \frac{45-54}{2} \right) - \left( \frac{-6-8}{3} \right)$$A = -\frac{9}{2} + \frac{14}{3} = \frac{-27+28}{6} = \frac{1}{6}$Thus, the area is $\frac{1}{6}$ square units.

Area of the region (in square units) bounded by the curve $y=x^2+4$ and the line $y=5x-2$ is

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