Solve ∫ (-1/2x^2+2,5) wrt x

$$\int ( - \frac { 1 } { 2 } x ^ { 2 } + 2,5 ) d x$$

Evaluate

$$-\frac{x^{3}}{6}+\frac{5x}{2}+С$$

Solution Steps

Integrate the sum term by term.

$$\int -\frac{x^{2}}{2}\mathrm{d}x+\int 2,5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{\int x^{2}\mathrm{d}x}{2}+\int 2,5\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -\frac{1}{2} times \frac{x^{3}}{3}.

$$-\frac{x^{3}}{6}+\int 2,5\mathrm{d}x$$

Find the integral of 2,5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{3}}{6}+\frac{5x}{2}$$

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.

$$-\frac{x^{3}}{6}+\frac{5x}{2}+С$$

Differentiate w.r.t. x

$$\frac{5-x^{2}}{2}$$