Solve ∫ (from 0 to pi div 2) of (3x+sin(x)) wrt x

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

Evaluate

$$\frac{3\pi ^{2}}{8}+1\approx 4.70110165$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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.

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

Simplify.

$$\frac{3\pi ^{2}}{8}+1$$