Solve ∫ (from 0 to 2) of (3x^2+4x+6y) wrt x

$$\int _ { 0 } ^ { 2 } ( 3 x ^ { 2 } + 4 x + 6 y ) d x$$

Evaluate

$$12y+16$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 4 times \frac{x^{2}}{2}.

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

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

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

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.

$$2^{3}+2\times 2^{2}+6y\times 2-\left(0^{3}+2\times 0^{2}+6y\times 0\right)$$

Simplify.

$$16+12y$$

Differentiate w.r.t. y

$$12$$