Solve ∫ (from 0 to 3) of (3x^2+6+cosx) wrt x

$$\int _ { 0 } ^ { 3 } ( 3 x ^ { 2 } + 6 + \cos x ) d x$$

Evaluate

$$\sin(3)+45\approx 45.141120008$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x^{2}+6+\cos(x)\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int 6\mathrm{d}x+\int \cos(x)\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x+\int 6\mathrm{d}x+\int \cos(x)\mathrm{d}x$$

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}.

$$x^{3}+\int 6\mathrm{d}x+\int \cos(x)\mathrm{d}x$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{3}+6x+\int \cos(x)\mathrm{d}x$$

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

$$x^{3}+6x+\sin(x)$$

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.

$$3^{3}+3\times 6+\sin(3)-\left(0^{3}+0\times 6+\sin(0)\right)$$

Simplify.

$$45+\sin(3)$$