Intergration Help!

**looi76** · February 28th 2008, 10:33 PM

Question:
Find the area enclosed between the curves $y = x^2 + 7$ and $y = 2x^2 + 3$ .

Attempt:

$[ y = x^2 + 7 ] \times 2$
$[ y = 2x^2 + 3 ]$

$2y = 2x^2 + 14$
$y = 2x^2 + 3$
$y = 11$

$11 = 2x^2 + 3$
$11 - 3 = 2x^2$
$8 = 2x^2$
$\frac{8}{2} = x^2$
$4 = x^2$
$x = \sqrt{4}$
$x = 2$

$y = 11 , x = 2$

$y = x^2 +7$

$= \frac{x^{2+1}}{2+1} +7x$

$= \frac{x^3}{3} + 7x$

$= (\frac{1}{3}x^3 + 7x)^2_0$

$= (\frac{1}{3}\times2^3 + 7\times2) - (\frac{1}{3}\times0^3 + 7\times0)$

$= 16\frac{2}{3} - 0$

$= 16\frac{2}{3}$

$y = 2x^2 + 3$

$= \frac{2x^{2+1}}{2+1} + 3x$

$= \frac{2x^3}{3} + 3x$

$= (\frac{2}{3}x^3 + 3x)^2_0$

$= (\frac{2}{3}\times2^3 + 3\times2) - (\frac{2}{3}\times0^3 + 3\times0)$

$= 11\frac{1}{3}$

Enclosed Area = $16\frac{2}{3} - 11\frac{1}{3}$
$= 5\frac{1}{3}$

Answer in the text book is $10\frac{2}{3}$

**Mathnasium** · February 28th 2008, 11:59 PM

First, find the points where the two functions intersect, which is to say you have to define the "ends" of the enclosed area. Both functions are equal to y, so set them equal to one another:

$x^2 + 7 = 2x^2 + 3$

Solving that gives $x= \pm 2$ - remember $\sqrt{4} = \pm 2$ - I think this is the step you missed.

In the interval (-2,2), $x^2 + 7$ has a greater value than $2x^2 + 3$ . In other words, $x^2 + 7$ will be the "top" function.

You're going to have to integrate with respect to x as x goes from -2 to 2, as those are the "ends" of the region of intersection. To find the area enclosed between two curves, you integrate $f(x) - g(x)$ where $f(x)$ is the "top" function and $g(x)$ is the "bottom" function.

So your integral is:

$\int^2_{-2} [(x^2 + 7) - (2x^2 + 3)] dx$

$= \int_{-2}^2 (-x^2 + 4) dx$

$= - \frac{1}{3}x^3 + 4x]^2_{-2}$

$= (- \frac{8}{3} + 8) - ( \frac{8}{3} - 8)$

$= 10 \frac{2}{3}$ .

Hope this helps!

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