Use the fundamental theorem of calculus to evaluate the definite integral
$\displaystyle\int_0^3 \frac{1}{\sqrt{1+x}}dx$
I dont get what they want here is it just to take the $F'(x) =\frac{1}{\sqrt{1+x}}$ where $x$ is $0$ and $3$ and find the mid point?
$\endgroup$2 Answers
$\begingroup$Hint: check that
$$\left(2\sqrt{1+x}\right)'=\frac{1}{\sqrt{1+x}}$$
$\endgroup$ 7 $\begingroup$The fundamental Theorem of Calculus says that, if $F'(x) = f(x)$, then
$$\int_a^b f(x) \, dx = F(b) - F(a)$$
That is, evaluate $F$ at the endpoints.
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