Solve ∫ (from - 2 to 6) of (x^5+3x^2+2x^4+3) wrt x=

$$\int _ { - 2 } ^ { 6 } ( x ^ { 5 } + 3 x ^ { 2 } + 2 x ^ { 4 } + 3 ) d x =$$

Evaluate

$$\frac{167048}{15}\approx 11136.533333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{5}+3x^{2}+2x^{4}+3\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{5}\mathrm{d}x+\int 3x^{2}\mathrm{d}x+\int 2x^{4}\mathrm{d}x+\int 3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{5}\mathrm{d}x+3\int x^{2}\mathrm{d}x+2\int x^{4}\mathrm{d}x+\int 3\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.

$$\frac{x^{6}}{6}+3\int x^{2}\mathrm{d}x+2\int x^{4}\mathrm{d}x+\int 3\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 3 times \frac{x^{3}}{3}.

$$\frac{x^{6}}{6}+x^{3}+2\int x^{4}\mathrm{d}x+\int 3\mathrm{d}x$$

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 2 times \frac{x^{5}}{5}.

$$\frac{x^{6}}{6}+x^{3}+\frac{2x^{5}}{5}+\int 3\mathrm{d}x$$

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

$$\frac{x^{6}}{6}+x^{3}+\frac{2x^{5}}{5}+3x$$

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.

$$\frac{6^{6}}{6}+6^{3}+\frac{2}{5}\times 6^{5}+3\times 6-\left(\frac{\left(-2\right)^{6}}{6}+\left(-2\right)^{3}+\frac{2}{5}\left(-2\right)^{5}+3\left(-2\right)\right)$$

Simplify.

$$\frac{167048}{15}$$