Question

Evaluate the integral...

Answer 1

\[\int\limits_{1}^{4}((1 + u)/\sqrt(u))du\]

Answer 2

$$\int\limits_{1}^{4}\frac{(1 + u)}{\sqrt{u}}du$$

Answer 3

yes

Answer 4

\[\Large \int_{1}^{4}\left(\frac{1+u}{\sqrt{u}}\right)du=\int_{1}^{4}\left(\frac{1}{\sqrt{u}}+\frac{u}{\sqrt{u}}\right)du\]
\[\Large \int_{1}^{4}\left(\frac{1+u}{\sqrt{u}}\right)du=\int_{1}^{4}\left(\frac{1}{u^{1/2}}\right)du+\int_{1}^{4}\left(\frac{u}{u^{1/2}}\right)du\]
I'll let you finish up

Answer 5

not sure about integrating from there, what would the antiderivative of the last part be?

Answer 6

Simplify each integrand and then use the rule

\[\Large \int x^n dx = \frac{x^{n+1}}{n+1} + C\]

n is a constant

Answer 7

\[\Large \int_{1}^{4}\left(\frac{1+u}{\sqrt{u}}\right)du=\int_{1}^{4}\left(\frac{1}{\sqrt{u}}+\frac{u}{\sqrt{u}}\right)du\]
\[\Large \int_{1}^{4}\left(\frac{1+u}{\sqrt{u}}\right)du=\int_{1}^{4}\left(\frac{1}{u^{1/2}}\right)du+\int_{1}^{4}\left(\frac{u}{u^{1/2}}\right)du\]
\[\Large \int_{1}^{4}\left(\frac{1+u}{\sqrt{u}}\right)du=\int_{1}^{4}\left(u^{-1/2}\right)du+\int_{1}^{4}\left(u^{1/2}\right)du\]
I'll let you finish up

Answer 8

ok, think I can, thanks jim

Answer 9

np