Solve ∫ (from 1 to x) of f(t)dt=x^3-x^2+ax

$$\int _ { 1 } ^ { x } f ( t ) d t = x ^ { 3 } - x ^ { 2 } + a x$$

Solve for a

$$\left\{\begin{matrix}a=\frac{\left(x-1\right)\left(f+fx-2x^{2}\right)}{2x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$x^{3}-x^{2}+ax=\int _{1}^{x}ft\mathrm{d}t$$

Subtract x^{3} from both sides.

$$-x^{2}+ax=\int _{1}^{x}ft\mathrm{d}t-x^{3}$$

Add x^{2} to both sides.

$$ax=\int _{1}^{x}ft\mathrm{d}t-x^{3}+x^{2}$$

The equation is in standard form.

$$xa=\frac{f\left(x^{2}-1\right)}{2}-x^{3}+x^{2}$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{\left(x-1\right)\left(f+fx-2x^{2}\right)}{2x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{\left(x-1\right)\left(f+fx-2x^{2}\right)}{2x}$$

Solve for f

$$\left\{\begin{matrix}f=-\frac{x\left(-x^{2}+x-a\right)}{t\left(x-1\right)}\text{, }&x\neq 1\text{ and }t\neq 0\\f\in \mathrm{R}\text{, }&\left(a=0\text{ and }x=1\right)\text{ or }\left(a\leq \frac{1}{4}\text{ and }t=0\text{ and }x=\frac{-\sqrt{1-4a}+1}{2}\right)\text{ or }\left(x=\frac{\sqrt{1-4a}+1}{2}\text{ and }a\leq \frac{1}{4}\text{ and }t=0\text{ and }a\neq 0\right)\text{ or }\left(t=0\text{ and }x=0\right)\end{matrix}\right.$$