int( absolute(x-1) + sqrt(9-x) )
Hint: you may interpret the integral as an area.
To find the value of the integral int(absolute(x-1) + sqrt(9-x)), we can interpret it as the area between the curve y = |x-1| + √(9-x) and the x-axis.
First, let's consider the function |x-1| + √(9-x). This function is not continuous for all values of x, so we need to split the integral into two parts.
1. For x ≤ 1:
When x is less than or equal to 1, the absolute value |x-1| evaluates as 1-x. So, the function becomes (1-x) + √(9-x).
2. For 1 < x ≤ 9:
When x is between 1 and 9, the absolute value |x-1| evaluates as x-1. So, the function becomes (x-1) + √(9-x).
Now, let's find the value of the integral for these two intervals and add them together.
1. For x ≤ 1:
int((1-x) + √(9-x)) dx from x = 0 to x = 1
To evaluate this integral, we can split it into two parts:
- int(1-x) dx from x = 0 to x = 1: This is a straight line with slope -1 and intercept 1. The integral evaluates as [-x^2/2 + x] from x = 0 to x = 1. Plugging in the limits gives -(1/2) + 1 = 1/2.
- int(√(9-x)) dx from x = 0 to x = 1: This is the area under the curve y = √(9-x) from x = 0 to x = 1. To evaluate this, we can use the antiderivative of √(9-x), which is [-2/3 (9-x)^(3/2)].
Plugging in the limits gives:
[-2/3 (9-1)^(3/2)] - [-2/3 (9-0)^(3/2)] = (2/3) - 6/3 = -4/3.
Summing up both parts:
(1/2) + (-4/3) = 3/6 - 8/6 = -5/6.
2. For 1 < x ≤ 9:
int((x-1) + √(9-x)) dx from x = 1 to x = 9
To evaluate this integral, we can split it into two parts:
- int(x-1) dx from x = 1 to x = 9: This is a straight line with slope 1 and intercept -1. The integral evaluates as [(x^2/2) - x] from x = 1 to x = 9. Plugging in the limits gives:
[(9^2/2) - 9] - [(1^2/2) - 1] = 81/2 - 9 - 1/2 + 1 = 71/2.
- int(√(9-x)) dx from x = 1 to x = 9: This is the area under the curve y = √(9-x) from x = 1 to x = 9. To evaluate this, we can use the antiderivative of √(9-x), which is [-2/3 (9-x)^(3/2)].
Plugging in the limits gives:
[-2/3 (9-9)^(3/2)] - [-2/3 (9-1)^(3/2)] = 0 - (-2/3 * 8) = 16/3.
Summing up both parts:
71/2 + 16/3 = 213/6 + 32/6 = 245/6.
Finally, we add the values obtained for both intervals:
-5/6 + 245/6 = 240/6 = 40.
Therefore, the value of the integral int(absolute(x-1) + sqrt(9-x)) is 40, which represents the total area between the curve and the x-axis.