Use the definite integral to find the area between the x-axis and f(x) over the indicated interval.
f(x) = -x^2 + 9 [0, 5]
I did this and got (10/3) . The answer key says (10/3) but my professor did it and got (98/3), which is one of this answer choices. My professor says to disregard the answer key and (10/3) is wrong. Answer keys, on rare occasions, have been wrong but I need to know which one is right.
f(x) = -x^2 + 9 [0, 5]
I did this and got (10/3) . The answer key says (10/3) but my professor did it and got (98/3), which is one of this answer choices. My professor says to disregard the answer key and (10/3) is wrong. Answer keys, on rare occasions, have been wrong but I need to know which one is right.
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I did this and got (10/3) too.
But on a second look I see the curve dips below the x axis at x = 3, so I think you have to integrate on interval 1 to 3 and add to positive area on interval 3 to 5 . . .
But on a second look I see the curve dips below the x axis at x = 3, so I think you have to integrate on interval 1 to 3 and add to positive area on interval 3 to 5 . . .
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f(x) = -x^2 + 9 and the x-axis (which of course is f(x) = 0).
So, first we have to find where these two meet, so set them equal to each other:
-x^2 + 9 = 0
Move the x^2:
x^2 = 9
By the square root, these functions meet in two places, x = -3 and x = 3. Since our bounds are [0, 5], we can disregard the x = -3 result, but now we have two different intervals, because the functions switch which one is 'higher' at x = 3. So, we need to determine which function is higher on [0, 3) and which is higher on (3, 5].
Pick a point in each interval, i will choose 1 and 4. Plug these into the original function:
f(1) = -(1)^2 + 9 = -1 + 9 = 8, and so it is higher than the x-axis on [0, 3)
f(4) = -(4)^2 + 9 = -16 + 9 = -7 and so it is lower than the x-axis on (3, 5]
From here, we create two integrals of the higher function minus the lower, and add them, to get:
(int(0, 3): -x^2 + 9) + (int(3, 5): -(-x^2 + 9))
Distribute the second minus sign:
(int(0, 3): -x^2 + 9) + (int(3, 5): x^2 - 9)
The integrals are easily done via the power rule:
((-1/3)x^3 + 9x | (0, 3)) + ((1/3)x^3 - 9x | (3, 5))
Plug in the bounds and solve:
(18 - 0) + (-10/3 - (-18))
18 + 44/3
98/3
Looks like your professor is correct!
So, first we have to find where these two meet, so set them equal to each other:
-x^2 + 9 = 0
Move the x^2:
x^2 = 9
By the square root, these functions meet in two places, x = -3 and x = 3. Since our bounds are [0, 5], we can disregard the x = -3 result, but now we have two different intervals, because the functions switch which one is 'higher' at x = 3. So, we need to determine which function is higher on [0, 3) and which is higher on (3, 5].
Pick a point in each interval, i will choose 1 and 4. Plug these into the original function:
f(1) = -(1)^2 + 9 = -1 + 9 = 8, and so it is higher than the x-axis on [0, 3)
f(4) = -(4)^2 + 9 = -16 + 9 = -7 and so it is lower than the x-axis on (3, 5]
From here, we create two integrals of the higher function minus the lower, and add them, to get:
(int(0, 3): -x^2 + 9) + (int(3, 5): -(-x^2 + 9))
Distribute the second minus sign:
(int(0, 3): -x^2 + 9) + (int(3, 5): x^2 - 9)
The integrals are easily done via the power rule:
((-1/3)x^3 + 9x | (0, 3)) + ((1/3)x^3 - 9x | (3, 5))
Plug in the bounds and solve:
(18 - 0) + (-10/3 - (-18))
18 + 44/3
98/3
Looks like your professor is correct!
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Ans
10/3
10/3