# Thread: U-Substitution and Area Between Curves

1. ## U-Substitution and Area Between Curves

1) Use the Substitution Formula to evaluate the integral.

I'm doing my best looking at my lecture notes and examples from the book to guide me into the procedure of substituting, but its not helping me. Could someone show me some light on how to get this started?

2) Find the area of the regions enclosed by the lines and curves given below.
y = x2 - 4x and y = 6x

so I graphed the two functions, the limits of integration for the left hand most region is a = 0 and b = 4 . to solve for the right hand limit, I set the equations equal to each other, and got b=10.

For $\displaystyle 0 \leq x \leq 4 : f(x) - g(x) = 6x - 0 = 6x$
For $\displaystyle 4 \leq x \leq 10 : f(x) - g(x) = 6x - (x^2 - 4x) = -x^2 +10x$

Add the Area of both regions:

Total Area = $\displaystyle \int^4_0 6x dx + \int^{10}_{4} (-x^2 + 10x) dx$

Integrate
$\displaystyle [3x^2]^4_0 + [\ \frac{-x^3}{3} + 5x^2 \ ]^{10}_{4}$

Solve:

$\displaystyle 3(4)^2 - 0 + ( \frac{-(4)^3}{3} + 5(4)^2) - ( \frac{-(10)^3}{3} + 5(10)^2)$

$\displaystyle 48 - 0 + \frac{-64}{3} + 80 - ( \frac{-1000}{3} - 500)$

$\displaystyle \frac{144}{3} - \frac{4}{3} + \frac{240}{3} + \frac{1000}{3} - \frac{1500}{3}$

$\displaystyle \frac{-120}{3} = -40$

Where did I screw up? It seems right to me...

2. $\displaystyle \int_{\pi}^{3\pi /2}\cot^5 \left ( \frac{\theta}{6}\right)\sec^2 \left ( \frac{\theta}{6}\right)~d\theta$

First use a substitution to get it in a form like
$\displaystyle \int \cot^5 x \sec^2 x~dx$

Then it is,
$\displaystyle \int \frac{\cos^5 x}{\sin^5 x}\cdot\frac{1}{\cos^2 x}~dx$

$\displaystyle \int \frac{\cos^3 x}{\sin^5 x}~dx$

$\displaystyle \int \frac{(\cos x)(\cos^2 x)}{\sin^5 x}~dx$

$\displaystyle \int \frac{(\cos x)(1 - \sin^2 x)}{\sin^5 x}~dx$

$\displaystyle \int \frac{\cos x}{\sin^5 x} - \frac{\cos x}{\sin^3 x}~dx$

Now apply the substitution $\displaystyle u = \sin x$ and it's easy from here.

3. y = x2 - 4x and y = 6x

so I graphed the two functions, the limits of integration for the left hand most region is a = 0 and b = 4 . to solve for the right hand limit, I set the equations equal to each other, and got b=10.

For $\displaystyle 0 \leq x \leq 4 : f(x) - g(x) = 6x - \color{red}0\color{black} = 6x$ <-- Why 0?
For $\displaystyle 4 \leq x \leq 10 : f(x) - g(x) = 6x - (x^2 - 4x) = -x^2 +10x$
Think again. Here's the graph:

4. Don't you divide up the graphs and use the x axis as the function of g(x)? Thats what I have in my lecture notes for the examples my professor did in class.

Maybe I am just confused entirely?

5. Originally Posted by casemeister
Don't you divide up the graphs and use the x axis as the function of g(x)? Thats what I have in my lecture notes for the examples my professor did in class.
You don't have to do this if you use f(x) - g(x). It's because the area under the x axis is negative. Just integrate from 0 to 10 f(x) - g(x).