Question

evaluate the intergral (4x-3sqrt(x)),0,4)

Answer 1

\[\int\limits_{0}^{4} (4x-3\sqrt{x})dx\]

Answer 2

First find the anti derivative. Use the power rule: ∫x^p dx=(x^(p+1))/(p+1)

You should get 2x^2-2x^(3/2). Next since it's a definite integral, plug in 4, plug in 0 and subtract the difference. So you get

(2(4)^2-2(4)^(3/2))-(2(0)^2-2(0)^(3/2))

Answer 3

can you explain the power rule
thats were im getting confused

Answer 4

It's the opposite of the power rule from derivatives. You add one to the exponent and you divide by the same number. So, for example let's do ∫ 3x^3 dx

The exponent is 3, so add 1 to it and you get 4. So your answer is:

(3x^4)/4

Answer 5

how did you get 2x^(3/2) in my problem?

Answer 6

sqrtx is the same as x^(1/2). The exponent is 1/2, add 1 and you get 3/2.

∫ 3sqrtx dx = (3x^(3/2))/(3/2) =(3x^(3/2)) * (2/3) = 2x^(3/2) (the 3s cancel out)

Answer 7

\[∫ 3\sqrt{x} dx = \frac{ 3x ^{3/2} }{ \frac{ 3 }{ 2 } } = 3x ^{3/2} * \frac{ 2 }{ 3 } = 2x ^{3/2}\]

Answer 8

oo i see