Evaluate the Definite Integral?

Question

Integral from 4 to 6 ((8x^2+8)/(sqrt(x))dx! I can't figure out this answer! the sqrt is for squar root!

Answer 1

Put x = t^2, so that dx = 2t dt. Since t sqrt(x), the new limits of integration are now 2 and sqrt(6). Substituting x = t^2 in the integral, we get

Int (2 to sqrt(6) ((8t^4 + 8)/t) * 2t dt = 16 Int (2 to sqrt(6) (t^4 + 1) dt =16 [t^5/5 + t] (2 to sqrt(6) = 16 [sqrt(6^5)/5 + sqrt(6) - 32/5 - 2]

Answer 2

You can split up the numerator and use exponents in place of the radical as follows:

⌡8x^2 / sqrt x + 8/ sqrt x dx

= ⌡8x^2 / x^.5 + 8/x^.5 dx
= ⌡8x^1.5 + 8x^-.5 dx
=(16/5) x^2.5 + 16x^.5 from 6 to 4
= (16/5)*6^2.5 + 16*6^.5 - (16/5*4^2.5 + 16*4^.5)
= (576/5) √6 + 16√6 - 512/5 - 32
= (496/5)*√6 - 42.4
= 200.59

check my arithmetic, I did this w/o a calc

Answer 3

first, factor out that 8.
Then, try to get it into an integrable form. There's no rule I know for integrating that, so let's try this:

8(x^2+1)/(sqrt x)=8(x^2/sqrt x)+8(1/sqrt x)=8x^(3/2)+8x^(-1/2).

Then integrate:8 int (x^(3/2)+x^(-1/2))=8((2/5)x^(5/2)+2x^(1/2))

Next you apply the fundamental theorem of calculus: using the formula you just figured out, find f(6)-f(4).

Check my work please!!! I think I did it right, but I could be wrong.

Answer 4

integral[4,6] ((8x^2+8)/sqrt(x)) dx ... let x = u^2, dx=2udu
=integral[2,sqrt(6)]((8u^4+8)/u) 2udu
=integral[2,sqrt(6)]((16u^4+16)) du
= [2,sqrt(6)] ( 16u^5/5+16u)
= (16/5)( 36sqrt(6) -32)+16(sqrt(6)-2)
= (16/5) (116sqrt(6)-42)
= (32/5) (58sqrt(6)-21)