How do you find the definite integral of an absolute function?
∫_0^3|2x-3|
lower limit= 0
upper limit = 3
Integrate it in two pieces.
For x<1.5, |2x -3| = -2x + 3
For x>or=1.5, |2x -3| = 2x + 3
I have no idea why you have the cube of zero in front of the |2x -3| What is the _0^3 supposed to be?
i don't know either lol, i tried to get it from word document and i copied and pasted it that way
"For x<1.5, |2x -3| = -2x + 3"
don't you mean .2x - 3?
No. Think about it. If x<1.5, 2x -3 is negative. The absolute value process reverses the signs of what is inside.
To find the definite integral of an absolute function, such as ∫_0^3|2x-3|, you need to break down the absolute function into separate intervals based on where the function changes its behavior.
In this case, the absolute function |2x-3| changes behavior when (2x-3) changes sign. So let's find the point at which (2x-3) changes sign:
2x - 3 = 0
2x = 3
x = 3/2
We can observe that (2x-3) changes sign at x = 3/2. So we divide the interval [0, 3] into two separate intervals: [0, 3/2] and [3/2, 3].
For the interval [0, 3/2]:
In this interval, |2x-3| simplifies to -(2x-3), since (2x-3) is negative. So the integral becomes -∫_0^(3/2)(2x-3)dx.
For the interval [3/2, 3]:
In this interval, |2x-3| simplifies to (2x-3), since (2x-3) is positive. So the integral becomes ∫_(3/2)^3(2x-3)dx.
Now, you can find the definite integrals separately for each interval using the power rule of integration. The integral of (2x-3) is (x^2 - 3x). So the definite integrals become:
For the interval [0, 3/2]:
-∫_0^(3/2)(2x-3)dx = -[x^2 - 3x]_0^(3/2).
For the interval [3/2, 3]:
∫_(3/2)^3(2x-3)dx = [x^2 - 3x]_(3/2)^3.
Now, you can substitute the upper and lower limits into these expressions and calculate the definite integrals for each interval:
For the interval [0, 3/2]:
-[x^2 - 3x]_0^(3/2) = -[(3/2)^2 - 3(3/2)] - [0^2 - 3(0)]
= -[(9/4) - (9/2)] - [0 - 0]
= -(9/4 - 18/4) - 0
= -(9/4 - 18/4)
= -(-9/4)
= 9/4
For the interval [3/2, 3]:
[x^2 - 3x]_(3/2)^3 = [(3)^2 - 3(3)] - [(3/2)^2 - 3(3/2)]
= [9 - 9] - [9/4 - 9/2]
= 0 - [9/4 - 18/4]
= 0 - (-9/4)
= 9/4
Finally, add the results from both intervals to get the total definite integral:
Definite integral of ∫_0^3|2x-3| = (9/4) + (9/4) = 18/4 = 9/2.