Question

Help with simple integration

Answer 1

\[\int\limits_{-1}^{h}\sqrt{x+1}dx= \frac{ 8 }{ 3 }\]

Answer 2

I need to find h but i don't know how

Answer 3

\[u=x+1\\
du=dx\]
\[\large\int_{-1}^h\sqrt{u}\;du=\left[\frac{2}{3}u^\frac{3}{2}\right]_{0}^{h+1}=\frac{8}{3}\]

Answer 4

How did you get 16/3 when the limits were [-1,3]? You used your calculator, didn't you?

Answer 5

What's the antiderivative of sqrt(x + 1)?

It's 2/3(x + 1)^(3/2) + C

Now from Fundamental Theorem of Calculus, we evaluate this antiderivative from -1 to h. This gives:\[\int\limits_{-1}^{h}\sqrt{x+1}dx=\left[ \frac{ 2 }{ 3 }(h+1)^{3/2}+C \right]-\left[ \frac{ 2 }{ 3 }(-1+1)^{3/2}+C \right]=\frac{ 8 }{ 3 }\]We notice that in that expansion, the 'C' will cancel out and we can solve for h.\[ \frac{ 2 }{ 3 }(h+1)^{3/2}- \frac{ 2 }{ 3 }(-1+1)^{3/2} =\frac{ 8 }{ 3 } \rightarrow \frac{ 2 }{ 3 }(h+1)^{3/2} =\frac{ 8 }{ 3 } \rightarrow (h+1)^{3/2} =4\]\[\rightarrow (h+1)^{3/2} =4 \rightarrow h=16^{1/3}-1 \rightarrow h =2\sqrt[3]{2}-1\]

@fey

Answer 6

@SithsAndGiggles
Used Integration by Substitution but if you haven't done that yet, then the above solution might appear to be more intuitive.