Solve ∫ (from 0 to 8) of (98x^5+43x^3)

$$\int _ { 0 } ^ { 8 } ( 98 x ^ { 5 } + 43 x ^ { 3 } )$$

Evaluate

$$\frac{12977152}{3}\approx 4325717.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 98x^{5}+43x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 98x^{5}\mathrm{d}x+\int 43x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$98\int x^{5}\mathrm{d}x+43\int x^{3}\mathrm{d}x$$

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}. Multiply 98 times \frac{x^{6}}{6}.

$$\frac{49x^{6}}{3}+43\int x^{3}\mathrm{d}x$$

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

$$\frac{49x^{6}}{3}+\frac{43x^{4}}{4}$$

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.

$$\frac{49}{3}\times 8^{6}+\frac{43}{4}\times 8^{4}-\left(\frac{49}{3}\times 0^{6}+\frac{43}{4}\times 0^{4}\right)$$

Simplify.

$$\frac{12977152}{3}$$