# Using integration to find the area between two lines

1. Jun 20, 2012

### Luke77

1. The problem statement, all variables and given/known data
Find the area under (x/3x) and above (x/3x^.5) between x=1 and x=4.

2. Relevant equations
-Area of a representative rectangle= ((x/3x)-(x/3x^.5))dx
-To integrate, raise the power of part of an expression and then divide the number in front by the raised exponent.

3. The attempt at a solution
I'm not the best at simplifying but here it goes:
4
∫(x/3x- x/3x^.5)dx
1

=[((x/2)^2) / (3/2x^2) - ((x/2^2) / (3/1.5)x^1.5] between 1 and 4

From there the answers are not reasonable (too low). Any help?

Last edited: Jun 20, 2012
2. Jun 20, 2012

### LCKurtz

Is that $\frac x {3x}$ or $(\frac x 3) x$? Ditto for the second one. And in either case you should simplify before integrating.
And that wouldn't be how you integrate fractions. Do the algebra first.

3. Jun 21, 2012

### dimension10

Like LCKurtz said, the way you write it is ambiguous. Write it in $\LaTeX$ and then just subtract one integral from the other. Either case, your integrals are wrong. What you did was make the terrible error that $\int\frac{f(x)}{g(x)}\mbox{d}x=\frac{\int f(x)\mbox{d}x}{\int g(x)\mbox{d}x}$ which is obviously wrong.

4. Jun 21, 2012

### HallsofIvy

Staff Emeritus
As the others have said, this does not really make sense. It would be more reasonable to write the first as simply 1/3 and the second as x^.5/3. If you cannot us LaTeX at least use parentheses to make clear what you mean.

Yes, the integral of $x^n$ is $x^{n+1}/(n+1)$ but you cannot just apply that to parts of an integral, in particular to the numerator and denominator of a fraction. Remember the "product rule" and "quotient rule" for differentiation? You did not just differentiate the parts of a product or quotient and cannot do that for integration.

$$\int_1^4 ((1/3)- (1/3)x^{.5})dx= (1/3)\int_1^4 (1- x^{.5})dx]$$
$$= \left[x- (1/1.5)x^{1.5}\right]_1^4$$

5. Jun 21, 2012

### Luke77

Thank you everyone so much! I forgot all about the product and quotient rule. I'm sorry for all of my mistakes, I just started learning calculus. Thanks again.

6. Jun 21, 2012

### Luke77

You're absolutely right but I think I was saying that I would integrate everything as x=4 and then subtract integrating it as x=1 to get my answer.