Question

using eight rectangles find the lower sum of this definite integral

Answer 1

\[\int\limits_{-1}^{3} x^3 dx\]

Answer 2

@sleepyhead314

Answer 3

Idk

Answer 4

its ok :)

Answer 5

@Luigi0210

Answer 6

Ok

u=2x+1 , du=2dx
du/2=dx
\[\text{}=\frac{1}{4}\int\limits (u-1) \sqrt{u} \, du\]

\[=\frac{1}{4}\int\limits \left(u^{3/2}-\sqrt{u}\right) \, du\]

\[\frac{1}{4}\int\limits u^{3/2} \, du-\frac{1}{4}\int\limits \sqrt{u} \, du\]

\[\text{}=\frac{1}{4}\int\limits u^{3/2} \, du-\frac{u^{3/2}}{6}\]

\[=\frac{u^{5/2}}{10}-\frac{u^{3/2}}{6}+\text{constant}\]

\[\text{}=\frac{1}{10} (2 x+1)^{5/2}-\frac{1}{6} (2 x+1)^{3/2}+\text{constant}\]

\[\text{}=\frac{1}{15} (2 x+1)^{3/2} (3 x-1)+\text{constant}\]

This might help.

Answer 7

one question too.
how would you also find the upper sum?