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

1. ## 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:

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My work:

I found the area separately, from left to right.

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

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

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

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

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

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

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

2. 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:

------

My work:

I found the area separately, from left to right.

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

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

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

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

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

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

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?

3. Thank you very much for your response, Chris L!

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