Solve ∫ (from - 1 to 2) of (2t+1)^2dt

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

Evaluate

$$21$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2t+1\right)^{2}.

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

Evaluate the indefinite integral first.

$$\int 4t^{2}+4t+1\mathrm{d}t$$

Integrate the sum term by term.

$$\int 4t^{2}\mathrm{d}t+\int 4t\mathrm{d}t+\int 1\mathrm{d}t$$

Factor out the constant in each of the terms.

$$4\int t^{2}\mathrm{d}t+4\int t\mathrm{d}t+\int 1\mathrm{d}t$$

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 4 times \frac{t^{3}}{3}.

$$\frac{4t^{3}}{3}+4\int t\mathrm{d}t+\int 1\mathrm{d}t$$

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}.

$$\frac{4t^{3}}{3}+2t^{2}+\int 1\mathrm{d}t$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}t=at.

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

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.

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

Simplify.

$$21$$