No the correct answer is t = 1201.09,

If x = h^1/2 - 5, then dx = [(1/2)h^-1/2]dh, or dh = 2dx(x+5).

The integral INT{[100pi/(5-h^1/2)]dx becomes -(200pi)Int{[(5/x)+1]dx} with limits 6 to 5.

The result is t = -200pi[5lnx + x] with those limits. Substituting in the limits one gets

t = 200pi[5ln6/5 + 1] = 1201.09.

are you using a Word (or similar) program, then scanning in the results in order to be able to use more advanced symbols? It is really annoying that Mybestanswer is so technologically retarded.

...Here is how I completed the integration.

I did not post this earlier since I did not know whether you were being graded on this work.

According to my calculations, the result I got is 1201,0986 and I have used the trapezoidal rule in order to calculated it.

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