Solve ∫ (from 0 to x) of (-6t-4t^3)dt

$$\int _ { 0 } ^ { x } ( - 6 t - 4 t ^ { 3 } ) d t$$

Evaluate

$$-x^{2}\left(x^{2}+3\right)$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -6t-4t^{3}\mathrm{d}t$$

Integrate the sum term by term.

$$\int -6t\mathrm{d}t+\int -4t^{3}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$-6\int t\mathrm{d}t-4\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}. Multiply -6 times \frac{t^{2}}{2}.

$$-3t^{2}-4\int t^{3}\mathrm{d}t$$

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

$$-3t^{2}-t^{4}$$

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.

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

Simplify.

$$-\left(3+x^{2}\right)x^{2}$$

Differentiate w.r.t. x

$$-4x^{3}-6x$$