Solve ∫ (from - 1 to 2) of (6t+5)(1-2t)dt

$$\int _ { - 1 } ^ { 2 } ( 6 t + 5 ) ( 1 - 2 t ) d t$$

Evaluate

$$-27$$

Solution Steps

Apply the distributive property by multiplying each term of 6t+5 by each term of 1-2t.

$$\int _{-1}^{2}6t-12t^{2}+5-10t\mathrm{d}t$$

Combine 6t and -10t to get -4t.

$$\int _{-1}^{2}-4t-12t^{2}+5\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int -4t-12t^{2}+5\mathrm{d}t$$

Integrate the sum term by term.

$$\int -4t\mathrm{d}t+\int -12t^{2}\mathrm{d}t+\int 5\mathrm{d}t$$

Factor out the constant in each of the terms.

$$-4\int t\mathrm{d}t-12\int t^{2}\mathrm{d}t+\int 5\mathrm{d}t$$

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 -4 times \frac{t^{2}}{2}.

$$-2t^{2}-12\int t^{2}\mathrm{d}t+\int 5\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}. Multiply -12 times \frac{t^{3}}{3}.

$$-2t^{2}-4t^{3}+\int 5\mathrm{d}t$$

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}t=at.

$$-2t^{2}-4t^{3}+5t$$

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.

$$5\times 2-2\times 2^{2}-4\times 2^{3}-\left(5\left(-1\right)-2\left(-1\right)^{2}-4\left(-1\right)^{3}\right)$$

Simplify.

$$-27$$