Problem in calculating the coefficients in a simple integral problem

Question

With $h(x)=\int_x^{x+3}a(t-3)(t+3)dt$, where $a>0$ is a constant, $$h'(x)=ax(x+6)-a(x-3)(x+3)=a(6x+9)$$ $$h(x)=\int a(6x+9)dx=a(3x^2+9x)+C$$

The book says $h(x)=a(3x^2+9x-18)$, and I don't understand how the constant part is calculated.

Another condition is that the minimum value of $h(x)$ is $-\frac{33}4$, so $h(-\frac32)=-\frac{33}4$, but I can only erase one constant with this, and I should get $h(x)$ with no unknown constant.

What is the hidden part to get $h(x)=a(3x^2+9x-18)$?

The final answer is $h(x)=x^2+3x-6$.

Answer 1

No constant part is calculated because $h(x)$ is a definite integral, just integrate the polynomial and you'll get that same expression.

Answer 2

Just integrate the polynomial $a(t-3)(t+3)$,

$h(x)=\int\limits_{x}^{x+3}{a({{t}^{2}}-9)dt}=a\left[ \frac{{{t}^{3}}}{3}-9t \right]_{x}^{x+3}$

$\Rightarrow h(x)=a\left[ \frac{{{(x+3)}^{3}}-{{x}^{3}}}{3}-9(x+3-x) \right]$

$\Rightarrow h(x)=a(3{{x}^{2}}+9x-18)$ ………… (1)

$=3a\left( {{\left( x+\frac{3}{2} \right)}^{2}}-\frac{33}{4} \right)$

Given that, then minimum value is $-\frac{33}{4}$.

This minimum value occurs at $x=-\frac{3}{2}$, that gives $3a=1$or $a=\frac{1}{3}$.

Therefore from equation (1), $h(x)={{x}^{2}}+3x-6$