Solve ∫ (from - 1 to 3) of (3x+2)^2 wrt x=

$$\int _ { - 1 } ^ { 3 } ( 3 x + 2 ) ^ { 2 } d x =$$

Evaluate

$$148$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$\int _{-1}^{3}9x^{2}+12x+4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 9x^{2}+12x+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9x^{2}\mathrm{d}x+\int 12x\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$9\int x^{2}\mathrm{d}x+12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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

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

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

$$3x^{3}+6x^{2}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$3x^{3}+6x^{2}+4x$$

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.

$$3\times 3^{3}+6\times 3^{2}+4\times 3-\left(3\left(-1\right)^{3}+6\left(-1\right)^{2}+4\left(-1\right)\right)$$

Simplify.

$$148$$