Intergration by substitution 2

• Jul 1st 2007, 11:52 AM
camherokid
Intergration by substitution 2
Find the area under the curve
• Jul 1st 2007, 12:03 PM
Jhevon
Quote:

Originally Posted by camherokid
Find the area under the curve

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

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

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

We proceed by substitution

Let $\displaystyle u = 2x + 1$

$\displaystyle \Rightarrow du = 2 ~dx$

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

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

and when $\displaystyle x = 0 \implies u = 1$ ..........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)

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

And i think you can take it from there
• Jul 1st 2007, 12:05 PM
red_dog
Let $\displaystyle \displaystyle I(x)=\int\sqrt{2x+1}dx$.
Substitution: $\displaystyle 2x+1=t\Rightarrow 2dx=dt$.
Then $\displaystyle \displaystyle I(t)=\frac{1}{2}\int\sqrt{t}dt=\frac{1}{2}\frac{t^ {\frac{3}{2}}}{\frac{3}{2}}+C$.
So $\displaystyle \displaystyle I(x)=\frac{1}{3}(2x+1)^{\frac{3}{2}}+C$