How do I find the integral of an absolute value?

Question

I'm given |2x+3| and I'm supposed to find the integral 2 over -4. I know the normal integral would be x^2 + 3x. But how does the absolute value make a difference?

Answer 1

Novangeli's answer is correct, your integral is the sul of the two terms Integral[-(2x+3) from -4 to -3/2 ] + Integral [2x+3 from -3/2 to 2 ].

On the contrary, what Greg says is "absolutely" not true. The absolute value of a continuous function change only the "smoothness" of its grafic representation, not its continuity, and a continuous function in a closed interval is always integrable!!!

Answer 2

The absolute value changes the sign of the function wherever it would otherwise be negative. For instance, if f(x) = 2x+3 then f(-2) = -1, but if f(x) = |2x+3|, f(-2) = 1.

In order to integrate over an absolute value, it is best to break the function into pieces, where the part inside the absolute value is always positive or always negative. 2x+3 is negative wherever x<-3/2, and nonnegative wherever x≥-3/2. So we can rewrite f(x) = |2x+3| as:

f(x) = {2x+3 if x≥-3/2, -(2x+3) if x<-3/2}

To integrate a piecewise function, you just break up the integral into several integrals, so that each integral only covers one piece. Thus:

[-4, 2]∫|2x+3| dx = [-4, -3/2]∫(-2x+3) dx + [-3/2, 2]∫2x+3 dx

Which you may now solve normally:

-(x²+3x)|[-4, -3/2] + (2x+3)|[-3/2, 2]
-(9/4 - 9/2 - 16 + 12) + (4+6 - 9/4 + 9/2)
-(-25/4) + (49/4)
49/4 + 25/4
74/4
37/2

Answer 3

The absolute value makes the function nonintegrable over the vertex of the graph. This occurs when the normal function would turn negative but the absolute value makes it positive. You have to perform 2 integrals. One goes from -4 to -3/2. The next goes from -3/2 to 2. You have to flip the sign of your function when you do the integral over the region that would normally be negative

Answer 4

Break it into two definite integrals:

-(2x+3) from -4 to -3/2
2x+3 from -3/2 to 2

Answer 5

the function passes through 0 at x=-3/2, so you need to integrate
. -3/2 .......... .. 2
∫-(2x + 3)dx + ∫(2x + 3)dx
-4..... ..... ..... -3/2