The value of int_{0}^{5} max {x^{2}, 6x-8} dx | vedclass

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(C) To evaluate $\int_{0}^{5} \max \{x^{2}, 6x-8\} dx$, we first find the intersection points of $y = x^{2}$ and $y = 6x-8$.Setting $x^{2} = 6x-8$, we

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(C) To evaluate $\int_{0}^{5} \max \{x^{2}, 6x-8\} dx$, we first find the intersection points of $y = x^{2}$ and $y = 6x-8$.Setting $x^{2} = 6x-8$, we get $x^{2}-6x+8 = 0$, which factors as $(x-2)(x-4) = 0$. Thus, the curves intersect at $x = 2$ and $x = 4$.For $x \in [0, 2]$, $x^{2} \ge 6x-8$.For $x \in [2, 4]$, $6x-8 \ge x^{2}$.For $x \in [4, 5]$, $x^{2} \ge 6x-8$.Therefore, the integral is split as:$\int_{0}^{2} x^{2} dx + \int_{2}^{4} (6x-8) dx + \int_{4}^{5} x^{2} dx$$= \left[ \frac{x^{3}}{3} \right]_{0}^{2} + \left[ 3x^{2}-8x \right]_{2}^{4} + \left[ \frac{x^{3}}{3} \right]_{4}^{5}$$= (\frac{8}{3} - 0) + ((48-32) - (12-16)) + (\frac{125}{3} - \frac{64}{3})$$= \frac{8}{3} + (16 - (-4)) + \frac{61}{3}$$= \frac{8}{3} + 20 + \frac{61}{3} = \frac{69}{3} + 20 = 23 + 20 = 43$.

The value of $\int_{0}^{5} \max \{x^{2}, 6x-8\} dx$ is

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