Solve ∫ (from 0 to 1) of (1-2x-3x^2) wrt x

$$\int_{ 0 }^{ 1 } (1-2x-3 { x }^{ 2 } ) d x$$

Evaluate

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Solution Steps

Evaluate the indefinite integral first.

$$\int 1-2x-3x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int -2x\mathrm{d}x+\int -3x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}x-2\int x\mathrm{d}x-3\int x^{2}\mathrm{d}x$$

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

$$x-2\int x\mathrm{d}x-3\int x^{2}\mathrm{d}x$$

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 -2 times \frac{x^{2}}{2}.

$$x-x^{2}-3\int x^{2}\mathrm{d}x$$

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 -3 times \frac{x^{3}}{3}.

$$x-x^{2}-x^{3}$$

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.

$$1-1^{2}-1^{3}-\left(0-0^{2}-0^{3}\right)$$

Simplify.

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