Solve ∫ (from 1 to 4) of (1/4x^4-x^3-3x^2+8x) wrt x

$$\int _ { 1 } ^ { 4 } ( \frac { 1 } { 4 } x ^ { 4 } - x ^ { 3 } - 3 x ^ { 2 } + 8 x ) d x$$

Evaluate

$$-\frac{78}{5}=-15.6$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{x^{4}}{4}-x^{3}-3x^{2}+8x\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{x^{4}}{4}\mathrm{d}x+\int -x^{3}\mathrm{d}x+\int -3x^{2}\mathrm{d}x+\int 8x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{\int x^{4}\mathrm{d}x}{4}-\int x^{3}\mathrm{d}x-3\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x$$

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 \frac{1}{4} times \frac{x^{5}}{5}.

$$\frac{x^{5}}{20}-\int x^{3}\mathrm{d}x-3\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x$$

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 -1 times \frac{x^{4}}{4}.

$$\frac{x^{5}}{20}-\frac{x^{4}}{4}-3\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x$$

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}. Multiply -3 times \frac{x^{3}}{3}.

$$\frac{x^{5}}{20}-\frac{x^{4}}{4}-x^{3}+8\int x\mathrm{d}x$$

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 8 times \frac{x^{2}}{2}.

$$\frac{x^{5}}{20}-\frac{x^{4}}{4}-x^{3}+4x^{2}$$

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.

$$\frac{4^{5}}{20}-\frac{4^{4}}{4}-4^{3}+4\times 4^{2}-\left(\frac{1^{5}}{20}-\frac{1^{4}}{4}-1^{3}+4\times 1^{2}\right)$$

Simplify.

$$-\frac{78}{5}$$