Solve ∫ 1/4(4-x) wrt x

$$\int \frac { 1 } { 4 } ( 4 - x ) d x$$

Evaluate

$$-\frac{x^{2}}{8}+x+С$$

Solution Steps

Use the distributive property to multiply \frac{1}{4} by 4-x.

$$\int \frac{1}{4}\times 4+\frac{1}{4}\left(-1\right)x\mathrm{d}x$$

Cancel out 4 and 4.

$$\int 1+\frac{1}{4}\left(-1\right)x\mathrm{d}x$$

Multiply \frac{1}{4} and -1 to get -\frac{1}{4}.

$$\int 1-\frac{1}{4}x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int -\frac{x}{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}x-\frac{\int x\mathrm{d}x}{4}$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x-\frac{\int x\mathrm{d}x}{4}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -\frac{1}{4} times \frac{x^{2}}{2}.

$$x-\frac{x^{2}}{8}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$x-\frac{x^{2}}{8}+С$$

Differentiate w.r.t. x

$$-\frac{x}{4}+1$$