Solve ∫ (7-2x)^3 wrt x

$$\int ( 7 - 2 x ) ^ { 3 } d x$$

Evaluate

$$-\frac{\left(7-2x\right)^{4}}{8}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(7-2x\right)^{3}.

$$\int 343-294x+84x^{2}-8x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 343\mathrm{d}x+\int -294x\mathrm{d}x+\int 84x^{2}\mathrm{d}x+\int -8x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 343\mathrm{d}x-294\int x\mathrm{d}x+84\int x^{2}\mathrm{d}x-8\int x^{3}\mathrm{d}x$$

Find the integral of 343 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$343x-294\int x\mathrm{d}x+84\int x^{2}\mathrm{d}x-8\int x^{3}\mathrm{d}x$$

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 -294 times \frac{x^{2}}{2}.

$$343x-147x^{2}+84\int x^{2}\mathrm{d}x-8\int x^{3}\mathrm{d}x$$

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

$$343x-147x^{2}+28x^{3}-8\int x^{3}\mathrm{d}x$$

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 -8 times \frac{x^{4}}{4}.

$$343x-147x^{2}+28x^{3}-2x^{4}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$343x-147x^{2}+28x^{3}-2x^{4}+С$$

Differentiate w.r.t. x

$$\left(7-2x\right)^{3}$$