Solve ∫ (from 0 to 2) of x^37-x^2 wrt x

$$\int_{ 0 }^{ 2 } { x }^{ 3 } 7- { x }^{ 2 } d x$$

Evaluate

$$\frac{76}{3}\approx 25.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 7x^{3}-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 7x^{3}\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$7\int x^{3}\mathrm{d}x-\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 7 times \frac{x^{4}}{4}.

$$\frac{7x^{4}}{4}-\int x^{2}\mathrm{d}x$$

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

$$\frac{7x^{4}}{4}-\frac{x^{3}}{3}$$

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.

$$\frac{7}{4}\times 2^{4}-\frac{2^{3}}{3}-\left(\frac{7}{4}\times 0^{4}-\frac{0^{3}}{3}\right)$$

Simplify.

$$\frac{76}{3}$$