Solve ∫ (from 0 to 6) of (0.55x^2)(1x)/2 wrt x

$$\int_{ 0 }^{ 6 } \frac{ (0.55 { x }^{ 2 } )(1x) }{ 2 } d x$$

Evaluate

$$89.1$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\int _{0}^{6}\frac{0.55x^{3}\times 1}{2}\mathrm{d}x$$

Multiply 0.55 and 1 to get 0.55.

$$\int _{0}^{6}\frac{0.55x^{3}}{2}\mathrm{d}x$$

Divide 0.55x^{3} by 2 to get 0.275x^{3}.

$$\int _{0}^{6}0.275x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int \frac{11x^{3}}{40}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$\frac{11\int x^{3}\mathrm{d}x}{40}$$

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}.

$$\frac{11x^{4}}{160}$$

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.

$$\frac{11}{160}\times 6^{4}-\frac{11}{160}\times 0^{4}$$

Simplify.

$$\frac{891}{10}$$