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

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

Evaluate

$$4305$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(4x-3\right)^{3}.

$$\int _{-2}^{5}64x^{3}-144x^{2}+108x-27\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 64x^{3}-144x^{2}+108x-27\mathrm{d}x$$

Integrate the sum term by term.

$$\int 64x^{3}\mathrm{d}x+\int -144x^{2}\mathrm{d}x+\int 108x\mathrm{d}x+\int -27\mathrm{d}x$$

Factor out the constant in each of the terms.

$$64\int x^{3}\mathrm{d}x-144\int x^{2}\mathrm{d}x+108\int x\mathrm{d}x+\int -27\mathrm{d}x$$

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

$$16x^{4}-144\int x^{2}\mathrm{d}x+108\int x\mathrm{d}x+\int -27\mathrm{d}x$$

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 -144 times \frac{x^{3}}{3}.

$$16x^{4}-48x^{3}+108\int x\mathrm{d}x+\int -27\mathrm{d}x$$

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

$$16x^{4}-48x^{3}+54x^{2}+\int -27\mathrm{d}x$$

Find the integral of -27 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$16x^{4}-48x^{3}+54x^{2}-27x$$

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.

$$16\times 5^{4}-48\times 5^{3}+54\times 5^{2}-27\times 5-\left(16\left(-2\right)^{4}-48\left(-2\right)^{3}+54\left(-2\right)^{2}-27\left(-2\right)\right)$$

Simplify.

$$4305$$