Solve ∫ (from 0 to 10) of x^3-41

$$\int _ { 0 } ^ { 10 } x ^ { 3 } - 41$$

Evaluate

$$2090$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}-41\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -41\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}+\int -41\mathrm{d}x$$

Find the integral of -41 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-41x$$

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.

$$\frac{10^{4}}{4}-41\times 10-\left(\frac{0^{4}}{4}-41\times 0\right)$$

Simplify.

$$2090$$