# Finding Area of Shaded Region - Thomas' Calculus P. 385 #39

• Mar 1st 2009, 10:12 AM
scherz0
Finding Area of Shaded Region - Thomas' Calculus P. 385 #39
Hello everyone,

Could someone please check my work to see where I've erred? My answer is slightly below the one given by my textbook.

Thank you!

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39. Find the total areas of the shaded region in:

http://img23.imageshack.us/img23/359...lusp38539h.png------

My work:

I found the area separately, from left to right.

Shaded part on left = $\displaystyle \int_{-2}^{-1}(-x + 2 - 4 + x^2) dx$

$\displaystyle = \int_{-2}^{-1}(x^2 - x - 2) dx$ $\displaystyle = \frac{x^3}{3} - \frac{x^2}{2} - 2x\bigg|_{-2}^{-1} = 11/6$

Shaded part in middle = $\displaystyle \int_{-2}^{2}(4 - x^2) dx - 7$

$\displaystyle = 4x - \frac{x^3}{3} - 2x\bigg|_{2}^{-2} - 6 = \frac{14}{3}$

Shaded part on right = $\displaystyle \int_{2}^{3}(-x + 2 - 4 + x^2) dx - \frac{1}{2}$

$\displaystyle = (\frac{x^3}{3} - \frac{x^2}{2} - 2x\bigg|_{2}^{3}) - \frac{1}{2} = \frac{11}{6} - \frac{1}{2} = \frac{4}{3}$

Therefore, total area = [tex] \frac{47}{6} [\math]. According to textbook, total area = 49/6 (hidden in white).
• Mar 1st 2009, 10:16 AM
Chris L T521
Quote:

Originally Posted by scherz0
Hello everyone,

Could someone please check my work to see where I've erred? My answer is slightly below the one given by my textbook.

Thank you!

---

39. Find the total areas of the shaded region in:

http://img23.imageshack.us/img23/359...lusp38539h.png------

My work:

I found the area separately, from left to right.

Shaded part on left = $\displaystyle \int_{-2}^{-1}(-x + 2 - 4 + x^2) dx$

$\displaystyle = \int_{-2}^{-1}(x^2 - x - 2) dx$ $\displaystyle = \frac{x^3}{3} - \frac{x^2}{2} - 2x\bigg|_{-2}^{-1} = 11/6$

Shaded part in middle = $\displaystyle \int_{-{\color{red}1}}^{2}(4 - x^2{\color{red}+x-2}) dx - 7$

$\displaystyle = 4x - \frac{x^3}{3} - 2x\bigg|_{2}^{-2} - 6 = \frac{14}{3}$

Shaded part on right = $\displaystyle \int_{2}^{3}(-x + 2 - 4 + x^2) dx {\color{blue}- \frac{1}{2}}$

$\displaystyle = (\frac{x^3}{3} - \frac{x^2}{2} - 2x\bigg|_{2}^{3}) {\color{blue}- \frac{1}{2}} = \frac{11}{6} {\color{blue}- \frac{1}{2}} = \frac{4}{3}$

Therefore, total area = [tex] \frac{47}{6} [\math]. According to textbook, total area = 49/6 (hidden in white).

You're missing the part in red, and your lower limit of integration for the middle part was wrong. And why is the bit in blue included?
• Mar 1st 2009, 10:28 AM
scherz0
Thank you very much for your response, Chris L!

Yes, the $\displaystyle -\frac{1}{2}$ was redundant.