Solve ∫ (from - 3 to 2) of (x^4+4x^3+12x^2) wrt x

$$\int _ { - 3 } ^ { 2 } ( x ^ { 4 } + 4 x ^ { 3 } + 12 x ^ { 2 } ) d x$$

Evaluate

$$130$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{4}+4x^{3}+12x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int 4x^{3}\mathrm{d}x+\int 12x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x+4\int x^{3}\mathrm{d}x+12\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}+4\int x^{3}\mathrm{d}x+12\int x^{2}\mathrm{d}x$$

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

$$\frac{x^{5}}{5}+x^{4}+12\int x^{2}\mathrm{d}x$$

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

$$\frac{x^{5}}{5}+x^{4}+4x^{3}$$

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.

$$\frac{2^{5}}{5}+2^{4}+4\times 2^{3}-\left(\frac{\left(-3\right)^{5}}{5}+\left(-3\right)^{4}+4\left(-3\right)^{3}\right)$$

Simplify.

$$130$$