Solve ∫ (from 1 to 6) of (4x^3-4x^4-3) wrt x

$$\int _ { 1 } ^ { 6 } ( 4 x ^ { 3 } - 4 x ^ { 4 } - 3 ) d x$$

Evaluate

$$-4940$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4x^{3}-4x^{4}-3\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{3}\mathrm{d}x+\int -4x^{4}\mathrm{d}x+\int -3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{3}\mathrm{d}x-4\int x^{4}\mathrm{d}x+\int -3\mathrm{d}x$$

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}. Multiply 4 times \frac{x^{4}}{4}.

$$x^{4}-4\int x^{4}\mathrm{d}x+\int -3\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -4 times \frac{x^{5}}{5}.

$$x^{4}-\frac{4x^{5}}{5}+\int -3\mathrm{d}x$$

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{4}-\frac{4x^{5}}{5}-3x$$

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.

$$6^{4}-\frac{4}{5}\times 6^{5}-3\times 6-\left(1^{4}-\frac{4}{5}\times 1^{5}-3\right)$$

Simplify.

$$-4940$$