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

$$\int_{ 1 }^{ 2 } 4- { x }^{ 2 } - \frac{ 1 }{ 4 } { x }^{ 3 } -2 d x$$

Evaluate

$$-\frac{61}{48}\approx -1.270833333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -1 times \frac{x^{3}}{3}.

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

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

$$4x-\frac{x^{3}}{3}-\frac{x^{4}}{16}+\int -2\mathrm{d}x$$

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$4x-\frac{x^{3}}{3}-\frac{x^{4}}{16}-2x$$

Simplify.

$$2x-\frac{x^{3}}{3}-\frac{x^{4}}{16}$$

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.

$$2\times 2-\frac{2^{3}}{3}-\frac{2^{4}}{16}-\left(2\times 1-\frac{1^{3}}{3}-\frac{1^{4}}{16}\right)$$

Simplify.

$$-\frac{61}{48}$$