Evaluate $\int_{0}^{5} \max ({{x}^{2},{6x-8}})\,dx$

Question

Evaluate $\displaystyle\int_{0}^{5} \max ({{x}^{2},{6x-8}})\,dx$

My attempts : $\max ({{x}^{2},{6x-8}})$= $6x -8 \,$ if $\,0≤{x}^{2}≤4$.

Now I take $\displaystyle\int_{0}^{2}{(6x-8)}\,dx +\int_{2}^{5}{{x}^{2}}\,dx = -4 + 39$.

I got the answer that is 35

Is my answer is correct or not? .

Please verify.

Thanks in advance

Answer 1

It's $$\int\limits_0^2x^2dx+\int\limits_4^5x^2dx+\int\limits_2^4(6x-8)dx=43$$ Because $x^2\geq6x-8$ for $x\in[0,2]\cup[4,5]$ and $6x-8\geq x^2$ for $x\in[2,4]$.

Answer 2

The best way to solve problems such as these is to draw a sketch of the curves involved. The straight line $y = 6x - 8$ intersects the parabola $y = x^2$ twice, at $x = 2$ and $x = 4$ with it now being a matter of finding which curve is larger (as we are asked to consider the maximum function) in the three regions: $0 \leqslant x \leqslant 2, 2 \leqslant x \leqslant 4$, and $4 \leqslant x \leqslant 5$.

So between 0 and 2 we have: $\text{max}\{x^2, 6x - 8\} = x^2$, between 2 and 4 we have: $\text{max}\{x^2,6x - 8\} = 6x - 8$, while between 4 and 5 we have: $\text{max}\{x^2,6x - 8\}$. So your definite integral becomes $$\int_0^5 \text{max}\{x^2, 6x - 8\} \, dx = \int_0^2 x^2 \, dx + \int_2^4 (6x - 8) \, dx + \int_4^5 x^2 \, dx,$$ and is something I am sure you can now readily evaluate.