Solve ∫ (from 0 to 2) of ((-2x^2+8-2.5x-3)+(2.5x+3+2x^2-8)) wrt x

$$\int_{ 0 }^{ 2 } ((-2 { x }^{ 2 } +8-2.5x-3)+(2.5x+3+2 { x }^{ 2 } -8)) d x$$

Evaluate

$$0$$

Solution Steps

Subtract 3 from 8 to get 5.

$$\int _{0}^{2}-2x^{2}+5-2.5x+2.5x+3+2x^{2}-8\mathrm{d}x$$

Combine -2.5x and 2.5x to get 0.

$$\int _{0}^{2}-2x^{2}+5+3+2x^{2}-8\mathrm{d}x$$

Add 5 and 3 to get 8.

$$\int _{0}^{2}-2x^{2}+8+2x^{2}-8\mathrm{d}x$$

Combine -2x^{2} and 2x^{2} to get 0.

$$\int _{0}^{2}8-8\mathrm{d}x$$

Subtract 8 from 8 to get 0.

$$\int _{0}^{2}0\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 0\mathrm{d}x$$

Find the integral of 0 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$0$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$0+0$$

Simplify.

$$0$$