Mathematics - Area Under Curves Question with Solution | TestHub

The given line is $4x+3y=12$, which can be written as $y=\frac{12-4x}{3}$.

The required area is bounded by $y=\frac{12-4x}{3}$, the X-axis ($y=0$), and the vertical lines $x=-1$ and $x=2$.

Since $y>0$ for all $x\in [-1,2]$, the area $A$ is given by the definite integral:

$A={\int }_{-1}^2\frac{12-4x}{3}dx$

$A=\frac{1}{3}{\left[12x-2x^2\right]}_{-1}^2$

$A=\frac{1}{3}\left((12(2)-2(2{)}^2)-(12(-1)-2(-1{)}^2)\right)$

$A=\frac{1}{3}\left((24-8)-(-12-2)\right)$

$A=\frac{1}{3}\left(16-(-14)\right)$

$A=\frac{1}{3}(30)=10$ sq units.

Answer: $10$