Solve ∫ (from 0 to 3 pi) of (2θ^2-sinx) wrt x

$$\int _ { 0 } ^ { 3 \pi } ( 2 \theta ^ { 2 } - \sin x ) d x$$

Evaluate

$$6\pi \theta ^{2}-2$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2\theta ^{2}-\sin(x)\mathrm{d}x$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

$$2\int \theta ^{2}\mathrm{d}x-\int \sin(x)\mathrm{d}x$$

Find the integral of \theta ^{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2\theta ^{2}x-\int \sin(x)\mathrm{d}x$$

Use \int \sin(\theta )\mathrm{d}\theta =-\cos(\theta ) from the table of common integrals to obtain the result. Multiply -1 times -\cos(x).

$$2\theta ^{2}x+\cos(x)$$

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.

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

Simplify.

$$6\theta ^{2}\pi -2$$

Differentiate w.r.t. θ

$$12\pi \theta$$