# Why is what I did wrong? (1 Viewer)

#### B1andB2

##### Well-Known Member
you only found a part of the shaded area (up until point B), but as you can see the area actually extends past that point

If you draw a line down from point C, you'll get a triangle. Shaded area is the total area of the triangle - the non-shaded area of the triangle.

The non-shaded area of the triangle is the same as the area from 0 to 2 which was part c)

Hence you minus (what they did)

#### Lith_30

##### Member
Your approach was kind of correct.

By solving for $\bg_white [{\frac{x^3}{3}-\frac{7x^2}{2}+10x]^5_2$ you would have gotten the area that is shaded in black.

Rather than doing that the answers give a more interesting method.

The area in blue represents $\bg_white \int_{0}^{2}(x^2-7x+10)dx$ since the graph is symmetrical we can also take the purple area to be the same. Looking at both the green and the purple areas we can see that they form a right angle triangle which has the area of $\bg_white \frac{1}{2}\times{5}\times {10}=25$. Then to find the green area we just simply subtract the purple area from the area of the triangle $\bg_white 25-\int_{0}^{2}(x^2-7x+10)dx=25-\frac{26}{3}=16\frac{1}{3} units^2$

#### Hivaclibtibcharkwa

##### Well-Known Member
Your approach was kind of correct.
View attachment 32753
By solving for $\bg_white [{\frac{x^3}{3}-\frac{7x^2}{2}+10x]^5_2$ you would have gotten the area that is shaded in black.

Rather than doing that the answers give a more interesting method.
View attachment 32754
The area in blue represents $\bg_white \int_{0}^{2}(x^2-7x+10)dx$ since the graph is symmetrical we can also take the purple area to be the same. Looking at both the green and the purple areas we can see that they form a right angle triangle which has the area of $\bg_white \frac{1}{2}\times{5}\times {10}=25$. Then to find the green area we just simply subtract the purple area from the area of the triangle $\bg_white 25-\int_{0}^{2}(x^2-7x+10)dx=25-\frac{26}{3}=16\frac{1}{3} units^2$
Thank you! I was having a hard time visualing everything, this helped

#### CM_Tutor

##### Moderator
Moderator
@Lith_30's illustration is covering what was meant for this question, IMO.

There is another approach, however. Draw in the vertical line $\bg_white x=5$ to divide the shaded region into a triangle and a region between the line $\bg_white y = 2x - 4$ and the curve. You then get:

\bg_white \begin{align*} \text{Shaded Area}\ &= \cfrac{1}{2} \times 3 \times 6 + \int_5^7 2x - 4 - \left(x^2 - 7x + 10\right)\ dx \\ &= 9 + \int_5^7 9x - x^2 - 14\ dx \\ &= 9 + \left[\cfrac{9x^2}{2} - \cfrac{x^3}{3} - 14x\right]_5^7 \\ &= 9 + \left[\left(\cfrac{9(7)^2}{2} - \cfrac{7^3}{3} - 14(7)\right) - \left(\cfrac{9(5)^2}{2} - \cfrac{5^3}{3} - 14(5)\right)\right] \\ &= 9 + \cfrac{9}{2}\left(7^2 - 5^2\right) - \cfrac{1}{3}\left(7^3 - 5^3\right) - 14(7 - 5) \\ &= 9 + \cfrac{9}{2} \times 2 \times 12 - \cfrac{1}{3} \times 2 \times \left(7^2 + 7 \times 5 + 5^2\right) - 14 \times 2 \\ &= 9 + 9 \times 12 - \cfrac{1}{3} \times 2 \times 109 - 14 \times 2 \\ &= 117 - \cfrac{218}{3} - 28 \\ &= 89 - 72\cfrac{2}{3} \\ &= 16\cfrac{1}{3}\ \text{sq. units} \end{align*}