Solve ∫ (from a to b) of 3x+2) wrt x=16quadtext{ve}quada+b=4

$$\int _ { a } ^ { b } 3 x + 2 ) d x = 16 \quad \text { ve } \quad a + b = 4$$

Evaluate

$$\frac{\left(b-a\right)\left(3a+3b+4\right)}{2}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x+2\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x\mathrm{d}x+\int 2\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x\mathrm{d}x+\int 2\mathrm{d}x$$

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

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{3}{2}b^{2}+2b-\left(\frac{3}{2}a^{2}+2a\right)$$

Simplify.

$$\frac{\left(4+3a+3b\right)\left(-a+b\right)}{2}$$

Differentiate w.r.t. a

$$-3a-2$$