Solve ∫ (from 0 to x) of f(t)dt=x^4/2+C

$$\int _ { 0 } ^ { x } f ( t ) d t = \frac { x ^ { 4 } } { 2 } + C$$

Solve for C

$$C=\frac{x^{2}\left(f-x^{2}\right)}{2}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2\int _{0}^{x}ft\mathrm{d}t=x^{4}+2C$$

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

$$x^{4}+2C=2\int _{0}^{x}ft\mathrm{d}t$$

Subtract x^{4} from both sides.

$$2C=2\int _{0}^{x}ft\mathrm{d}t-x^{4}$$

The equation is in standard form.

$$2C=fx^{2}-x^{4}$$

Divide both sides by 2.

$$\frac{2C}{2}=\frac{x^{2}\left(f-x^{2}\right)}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$C=\frac{x^{2}\left(f-x^{2}\right)}{2}$$

Solve for f

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