Question

Please help, evaluate the integral: question will be posted.

Answer 1

\[\int\limits_{1}^{4} \frac{ dt }{ t \sqrt{t} }\]

Answer 2

Please help using the fundamental theorem of calculus

Answer 3

Remember that \[
\sqrt{t} = t^{1/2}
\]And: \[
t\cdot t^{1/2} = t^{3/2}
\]AND: \[
\int x^ndx = \frac{x^{n+1}}{n+1}+C
\]

Answer 4

\[t* \sqrt{t} = t^1 * t^{1/2} = t^{3/2}\]

Answer 5

Also \[ \large
\frac{dt}{t^{n}} = t^{-n}dt
\]

Answer 6

So it can be rewritten as \[\int\limits_{1}^{4} t ^{3/2}dt\]

Answer 7

Fundamental Theorem (Part II) just says: \[
F'(t) = f(t) \implies \int_a^b f(t)dt = F(b) - F(a)
\]

Answer 8

@Brittni0605 Should be: \[
\int\limits_1^4 t^{-3/2}dt
\]Remember it was initially in the denominator.

Answer 9

Ok, so then I find the antiderivative and evaluate at the bounds?

Answer 10

Yes

Answer 11

Ok, I got an answer of 1. Is that correct?