Solve ∫ (from 3 to 5) of (4x^3-6x^2+5x+3) wrt x

$$\int _ { 3 } ^ { 5 } ( 4 x ^ { 3 } - 6 x ^ { 2 } + 5 x + 3 ) d x$$

Evaluate

$$394$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4x^{3}-6x^{2}+5x+3\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{3}\mathrm{d}x+\int -6x^{2}\mathrm{d}x+\int 5x\mathrm{d}x+\int 3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{3}\mathrm{d}x-6\int x^{2}\mathrm{d}x+5\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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

$$x^{4}-6\int x^{2}\mathrm{d}x+5\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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 -6 times \frac{x^{3}}{3}.

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

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

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

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{4}-2x^{3}+\frac{5x^{2}}{2}+3x$$

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.

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

Simplify.

$$394$$