Solve ∫ (x-2)(2x+5)-7(-10+4)

$$\int ( x - 2 ) ( 2 x + 5 ) - 7 ( - 10 + 4 )$$

Evaluate

$$\frac{2x^{3}}{3}+\frac{x^{2}}{2}+32x+С$$

Solution Steps

Apply the distributive property by multiplying each term of x-2 by each term of 2x+5.

$$\int 2x^{2}+5x-4x-10-7\left(-10+4\right)\mathrm{d}x$$

Combine 5x and -4x to get x.

$$\int 2x^{2}+x-10-7\left(-10+4\right)\mathrm{d}x$$

Add -10 and 4 to get -6.

$$\int 2x^{2}+x-10-7\left(-6\right)\mathrm{d}x$$

Multiply 7 and -6 to get -42.

$$\int 2x^{2}+x-10-\left(-42\right)\mathrm{d}x$$

The opposite of -42 is 42.

$$\int 2x^{2}+x-10+42\mathrm{d}x$$

Add -10 and 42 to get 32.

$$\int 2x^{2}+x+32\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 32\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 32\mathrm{d}x$$

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

$$\frac{2x^{3}}{3}+\int x\mathrm{d}x+\int 32\mathrm{d}x$$

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}.

$$\frac{2x^{3}}{3}+\frac{x^{2}}{2}+\int 32\mathrm{d}x$$

Find the integral of 32 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}+\frac{x^{2}}{2}+32x$$

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.

$$\frac{2x^{3}}{3}+\frac{x^{2}}{2}+32x+С$$

Differentiate w.r.t. x

$$2x^{2}+x+32$$