Mathematics - Area Under Curves Question with Solution | TestHub

Given the curve $y=x^3+ax$ with $a>0$, the roots of $y=0$ are given by $x(x^2+a)=0$. Since $a>0$, the only real root is $x=0$.

For $x\in [-2,0]$, $y\le 0$, and for $x\in [0,1]$, $y\ge 0$.

The required area $A$ is given by:

$A={\int }_{-2}^0-(x^3+ax)dx+{\int }_0^1(x^3+ax)dx$

$A=-{\left[\frac{x^4}{4}+\frac{a{x}^2}{2}\right]}_{-2}^0+{\left[\frac{x^4}{4}+\frac{a{x}^2}{2}\right]}_0^1$

$A=-\left(0-\left(\frac{16}{4}+\frac{4a}{2}\right)\right)+\left(\frac{1}{4}+\frac{a}{2}-0\right)$

$A=(4+2a)+\left(\frac{1}{4}+\frac{a}{2}\right)$

$A=\frac{17}{4}+\frac{5a}{2}$

Given that the area is $\frac{37}{4}$ square units:

$\frac{17}{4}+\frac{5a}{2}=\frac{37}{4}$

$\frac{5a}{2}=\frac{20}{4}$

$\frac{5a}{2}=5\Rightarrow a=2$

Answer: $a=2$