Solve ∫ (from 0 to 1) of (x^18+18^x) wrt x

$$\int _ { 0 } ^ { 1 } ( x ^ { 18 } + 18 ^ { x } ) d x$$

Evaluate

$$\frac{17}{\ln(18)}+\frac{1}{19}\approx 5.934227935$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{18}+18^{x}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{18}\mathrm{d}x+\int 18^{x}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{18}\mathrm{d}x with \frac{x^{19}}{19}.

$$\frac{x^{19}}{19}+\int 18^{x}\mathrm{d}x$$

Use \int x^{k}\mathrm{d}k=\frac{x^{k}}{\ln(x)} from the table of common integrals to obtain the result.

$$\frac{x^{19}}{19}+\frac{18^{x}}{\ln(18)}$$

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.

$$\frac{1^{19}}{19}+18^{1}\ln(18)^{-1}-\left(\frac{0^{19}}{19}+18^{0}\ln(18)^{-1}\right)$$

Simplify.

$$\frac{1}{19}+\frac{17}{\ln(18)}$$