Math Question Q13 (1 Viewer)

lolzlolz · May 1, 2022

Does anyone know how to do this???

lolzlolz · May 1, 2022

lolzlolz said:
Does anyone know how to do this???

it's q13

jimmysmith560 · May 2, 2022

Here is what I did:

Using $\text{Area}=\int _a^bf\left(y\right)dy$

$=\int _{-3}^01-y\:dy$

$=\left[y-\frac{y^2}{2}\right]^0_{-3}$

$=0-\frac{\left(-3\right)^2}{2}$

$=-\frac{9}{2}$

The integral is negative given that the area between $y=4$ and $y=1-x$ is in the second quadrant (i.e. to the left of the y-axis).

$\therefore \text{Area}=\left|-\frac{9}{2}\right|$

$=\frac{9}{2}\:\text{i.e.}\:4.5\:\text{units}^2,\:\text{matching the answer provided.}$

I hope this helps!

VBN2470 · May 8, 2022

Alternatively, note that the area of the region enclosed between the lines $y = 4$ and $y = 1 - x$ in the second quadrant is simply the area of an isosceles right triangle with side length $3$ (why?). This will give the desired result.

Math Question Q13 (1 Viewer)

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