The area of the region (in sq. units) enclosed by the curve $y=x^3-19 x+30$ and the $x$-axis is

Question

Aicharya · Accepted Answer

Given the curve, $y=x^3-19 x+30$ $\Rightarrow y=(x+5)(x-2)(x-3)$

$=\int_{-5}^2 y d x-\int_2^3 y d x$ $=\int_{-5}^2\left(x^3-19 x+30\right) d x-\int_2^3\left(x^3-19 x+30\right) d x$ $=\left(\frac{x^4}{4}-\frac{19 x^2}{2}+30 x\right)_{-5}^2-\left(\frac{x^4}{4}-\frac{19 x^2}{2}+30 x\right)_2^3$ $=(4-38+60)-\left(\frac{625}{4}-\frac{475}{2}-150\right)$ $\begin{aligned} & -\left(\frac{81}{4}-\frac{171}{2}+90\right)+(4-38+60)=\frac{1029}{4}+\frac{5}{4} \\ & =\frac{517}{2} \text { sq. units }\end{aligned}$

The area of the region (in sq. units) enclosed by the curve $y=x^3-19 x+30$ and the $x$-axis is

Solution