Solve ∫ (from 0.2 to 0.4) of 1/22125x^2 wrt x

$$\int _ { 0.2 } ^ { 0.4 } \frac { 1 } { 2 } 2125 x ^ { 2 } d x$$

Evaluate

$$\frac{119}{6}\approx 19.833333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{2125x^{2}}{2}\mathrm{d}x$$

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

$$\frac{2125}{2}\int x^{2}\mathrm{d}x$$

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}.

$$\frac{2125}{2}\times \frac{x^{3}}{3}$$

Simplify.

$$\frac{2125x^{3}}{6}$$

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.

$$\frac{2125}{6}\times 0.4^{3}-\frac{2125}{6}\times 0.2^{3}$$

Simplify.

$$\frac{119}{6}$$