Integral problem

[Chris283]

How do I solve this?

\(\displaystyle 17)\, \, \int_1^4\,\mbox{ max }\, \left(\, \dfrac{x^2}{2},\, \sqrt{2x\,}\, \right)\, dx;\)

 

[stapel]

How do I solve this?

\(\displaystyle 17)\, \, \int_1^4\,\mbox{ max }\, \left(\, \dfrac{x^2}{2},\, \sqrt{2x\,}\, \right)\, dx;\)

Looking at the graphs of f(x) = x²/2 and g(x) = sqrt[2x] over the interval x = 1 to x = 4, where do the lines intersect? Before that intersection point, which line is "higher" (that is, which of the functions is greater)? After that intersection point, which is higher?

Split the integral into two at that intersection point (call it "x = c"), so you have one integral from x = 1 to x = c and another from x = c to x = 4. On each interval, integrate the function which was higher on that interval. Then sum the results to get the answer.

If you get stuck, please reply showing your efforts so far. Thank you!

 

[Chris283]

ok, iwve trid to solve this and it seems that i have a caculation mistake- and I dont know where.

can you help?

I think the problem is at wrong caculation of sqrt(2x) somewhehr...

 

[Ishuda]

From what I can see you have the proper integrals
\(\displaystyle \int_1^2 \sqrt{2x} dx + \int_2^4 \frac{x^2}{2} dx\)
=\(\displaystyle \sqrt{2}\int_1^2 \sqrt{x} dx + \frac{1}{2} \int_2^4 x^2 dx\)
=\(\displaystyle \sqrt{2}\, \dfrac{2}{3}\, x^{\frac{3}{2}}|_1^2\, +\, \dfrac{1}{2} \dfrac{1}{3} x^3|_2^4\)
=\(\displaystyle \dfrac{2\sqrt{2}}{3}\, x^{\frac{3}{2}}|_1^2\, +\, \dfrac{1}{6}\, x^3|_2^4\)
=\(\displaystyle \dfrac{2\sqrt{2}}{3}\, (2\sqrt{2} - 1)\, +\, \dfrac{1}{6}\, (64\, -\, 8)\)
=\(\displaystyle \dfrac{1}{3}\, (8\, -\, 2\sqrt{2})\, +\, \dfrac{1}{6}\, (56)\)
=\(\displaystyle \dfrac{1}{3}\, (8\, -\, 2\sqrt{2})\, +\, \dfrac{1}{3}\, (28)\)
=\(\displaystyle -\, 2\sqrt{2})\, +\, \dfrac{1}{3}\, (8 + 28)\)
=\(\displaystyle \dfrac{1}{3}\, (36)\, -\, \dfrac{2\,\sqrt{2}}{3}\)
=\(\displaystyle 12\, -\, \dfrac{2\,\sqrt{2}}{3}\)
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