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

$$\int _ { 0 } ^ { 1 } ( 2 x ^ { 2 } - 3 \cos x + 1 ) d x$$

Evaluate

$$\frac{5-9\sin(1)}{3}\approx -0.857746288$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$\frac{2x^{3}}{3}-3\int \cos(x)\mathrm{d}x+\int 1\mathrm{d}x$$

Use \int \cos(x)\mathrm{d}x=\sin(x) from the table of common integrals to obtain the result.

$$\frac{2x^{3}}{3}-3\sin(x)+\int 1\mathrm{d}x$$

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

$$\frac{2x^{3}}{3}-3\sin(x)+x$$

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.

$$\frac{2}{3}\times 1^{3}-3\sin(1)+1-\left(\frac{2}{3}\times 0^{3}-3\sin(0)\right)$$

Simplify.

$$\frac{1}{3}\left(5-9\sin(1)\right)$$