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

$$\int_{ -5 }^{ 3 } 3 { x }^{ 3 } +x-4 d x$$

Evaluate

$$-448$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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

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

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.

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

Simplify.

$$-448$$