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

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

Evaluate

$$172$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 6 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}+2x^{3}$$

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.

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

Simplify.

$$172$$