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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?

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Hint: check that

$$\left(2\sqrt{1+x}\right)'=\frac{1}{\sqrt{1+x}}$$

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  • $\begingroup$ So it would end up F(b)-F(a) = -4--2 = -2?? my book tells me it should be 2 :( $\endgroup$ – 1ftw1 Apr 11 '13 at 10:29
  • $\begingroup$ His Integrand is $\frac{1}{\sqrt{1+x}}$, not what you have written. $\endgroup$ – Gautam Shenoy Apr 11 '13 at 10:37
  • $\begingroup$ Someone edited it after i posted so i just put it back sorry $\endgroup$ – 1ftw1 Apr 11 '13 at 10:40
  • $\begingroup$ He did what!! My apologies if this is the case. In the original integrand, the integrand is not defined from 1 to 3. Now it is fine. $\endgroup$ – Gautam Shenoy Apr 11 '13 at 10:41
  • $\begingroup$ sorry someone edited it its all right now sorry for the time wasting $\endgroup$ – 1ftw1 Apr 11 '13 at 10:42
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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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  • $\begingroup$ So it would end up F(b)-F(a) = -4--2 = -2?? my book tells me it should be 2 :( $\endgroup$ – 1ftw1 Apr 11 '13 at 10:34

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