Question

If the normal to the curve $y(x)={\int }_0^x\left(2t^2-15t+10\right)dt$ at a point $(a,b)$ is parallel to the line $x+3y=-5,a>1$, then the value of $|a+6b|$ is equal to ________.

Answer 1

Answer: 406

$y(x)={\int }_0^x\left(2t^2-15t+10\right)dt$
${y^{\prime }(x)]}_{x=a}={\left[2x^2-15x+10\right]}_a=2a^2-15a+10$
Slope of normal $=-\frac{1}{3}$
$\Rightarrow 2a^2-15a+10=3\Rightarrow a=7$
$\&a=\frac{1}{2}$ (rejected)
$b=y(7)={\int }_0^7\left(2t^2-15t+10\right)dt$
$={\left[\frac{2t^3}{3}-\frac{15t^2}{2}+10t\right]}_0^7$
$\Rightarrow 6b=4\times 7^3-45\times 49+60\times 7$
$|a+6b|=406$