Solve ∫ (from 0 to frac { pi) of {2}}(x^2+sinx) wrt x

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } ( x ^ { 2 } + \sin x ) d x$$

Evaluate

$$\frac{\pi ^{3}}{24}+1\approx 2.291928195$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

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

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

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

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.

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

Simplify.

$$\frac{\pi ^{3}}{24}+1$$