Solve g ( x ) = ∫ (from - 12 to x) of (4t-7)dt

$$g ( x ) = \int _ { - 12 } ^ { x } ( 4 t - 7 ) d t$$

Evaluate

$$\left(2x-31\right)\left(x+12\right)$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4t-7\mathrm{d}t$$

Integrate the sum term by term.

$$\int 4t\mathrm{d}t+\int -7\mathrm{d}t$$

Factor out the constant in each of the terms.

$$4\int t\mathrm{d}t+\int -7\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}. Multiply 4 times \frac{t^{2}}{2}.

$$2t^{2}+\int -7\mathrm{d}t$$

Find the integral of -7 using the table of common integrals rule \int a\mathrm{d}t=at.

$$2t^{2}-7t$$

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.

$$2x^{2}-7x-\left(2\left(-12\right)^{2}-7\left(-12\right)\right)$$

Simplify.

$$\left(-31+2x\right)\left(12+x\right)$$

Differentiate w.r.t. x

$$4x-7$$