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

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

Evaluate

$$\frac{7-2\cos(1)}{2}\approx 2.959697694$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$x^{3}+3\int x\mathrm{d}x+\int \sin(x)\mathrm{d}x$$

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

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

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

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

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.

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

Simplify.

$$\frac{1}{2}\left(7-2\cos(1)\right)$$