Intergration by substitution 2

**camherokid** · July 1st 2007, 11:52 AM

Find the area under the curve

**Jhevon** · July 1st 2007, 12:03 PM

Originally Posted by camherokid

Find the area under the curve

$y = \sqrt {2x + 1}$

The area under the curve between the limits $0 \leq x \leq 1$ is given by:

$A = \int_{0}^{1} \sqrt {2x + 1}~dx$

We proceed by substitution

Let $u = 2x + 1$

$\Rightarrow du = 2 ~dx$

$\Rightarrow \frac {1}{2} du = dx$

Also, when $x = 1 \implies u = 3$

and when $x = 0 \implies u = 1$ ..........we can use these to change the limits (I wouldn't, but that seems to be the preferred method around here, i always back-substitute to get the integral as a function of x and plug in the old limits)

$\Rightarrow A = \frac {1}{2} \int_{1}^{3} \sqrt {u}~du$

And i think you can take it from there

**red_dog** · July 1st 2007, 12:05 PM

Let $\displaystyle I(x)=\int\sqrt{2x+1}dx$ .
Substitution: $2x+1=t\Rightarrow 2dx=dt$ .
Then $\displaystyle I(t)=\frac{1}{2}\int\sqrt{t}dt=\frac{1}{2}\frac{t^ {\frac{3}{2}}}{\frac{3}{2}}+C$ .
So $\displaystyle I(x)=\frac{1}{3}(2x+1)^{\frac{3}{2}}+C$

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