Let f(x) =
0 if x < 4
5 if -4 ≤ x < -1
-4 if -1 ≤ x < 5
0 if x ≥ 5
g(x) = ∫ [upper limit x, lower limit -4] f(t)dt
Determine g(-5), g(−3), g(0), and g(6).
Also, the absolute maximum of g(x) occurs when x = ? and and is the value ?
draw the graph of f(t). It is just a pair of horizontal lines,
one at f=5 and one at f = -4
The integral is just the algebraic sum of the areas of rectangles:
g(-5) = ∫[-4,-5] = 0
g(-3) = ∫[-4,-3] = 5*1 = 5
g(0) = ∫[-4,0] = ∫[-4,-1] + ∫[-1,0] = 5*3 - 4*1 = 11
g(6) = ∫[-4,6] = ∫[-4,-1] + ∫[-1,5] + ∫[5,6] = 5*3 - 4*6 + 0*1 = -9
Oh boy, let's solve this integral puzzle! But first, let's have some fun!
Q: What did the calculus student say to his teacher when she asked him why he always forgets the constant of integration?
A: "Well, it must have slipped my mind!"
Alright, let's get back to business. We need to find the values of g(x) for different x values. Let's start with g(-5). To find this value, we need to integrate f(t) from the upper limit (-5) to the lower limit (-4).
Now, f(t) is 0 for t < 4, which means g(t) is also 0. So, the integral of f(t) from -5 to -4 will be 0 as well. Therefore, g(-5) = 0.
Moving on to g(-3), we still have the same situation where f(t) is 0 for t < 4. So, g(t) = 0 for -4 ≤ t < -1. But when t = -3, f(t) is 5. So, the integral of f(t) from -4 to -3 will be 5. Therefore, g(-3) = 5.
Now, let's tackle g(0). According to the function f(x), it's 0 if x ≥ 5. Since our upper limit is 0, the integral of f(t) from -4 to 0 will include all the intervals where f(t) is non-zero. So, we need to integrate f(t) from -1 to 0 and from -4 to -1. In both intervals, f(t) is -4. So, the integral will be -4 times the width of the interval, which is 3 (from -4 to -1). Therefore, g(0) = -4 * 3 = -12.
Lastly, let's determine g(6). Since f(x) is 0 for x ≥ 5, the integral of f(t) from -4 to 6 will only include the intervals where f(t) is non-zero. Those intervals are from -1 to 5, where f(t) is -4. So, the integral will be -4 times the width of the interval, which is 6 (from -4 to 5). Therefore, g(6) = -4 * 6 = -24.
To find the absolute maximum of g(x), we need to check all the values we found. So far, we have g(-5) = 0, g(-3) = 5, g(0) = -12, and g(6) = -24. The absolute maximum occurs when x = -3, and the value of g(x) at that point is 5.
And there you have it! I hope you enjoyed clowning around with integrals. If you have any more questions or need a good laugh, feel free to ask!
To compute g(x) for different values, we need to integrate the function f(t) within the specified limits. Let's break it down step-by-step:
1. g(x) = ∫[upper limit x, lower limit -4] f(t)dt
This equation represents the integral of f(t) with the upper limit being x and the lower limit being -4.
2. We'll calculate g(-5):
Since -5 < -4 ≤ x < -1, this falls under the second part of the function f(x) where f(x) = 5.
Thus, g(-5) = ∫[-5, -4] f(t)dt = ∫[-5, -4] 5 dt.
3. Evaluate ∫[-5, -4] 5 dt:
Since the function is constant, the integral simplifies to the product of the integrand (5) and the width of the interval ([-5, -4]), which is 1.
Therefore, ∫[-5, -4] 5 dt = 5 * 1 = 5.
4. g(-5) = 5.
5. Calculate g(-3):
Since -4 ≤ -3 < -1, this lies within the third part of f(x) where f(x) = -4.
Therefore, g(-3) = ∫[-3, -4] f(t)dt = ∫[-3, -4] -4 dt.
6. Evaluate ∫[-3, -4] -4 dt:
Similar to the previous step, the integral simplifies to the product of the integrand (-4) and the width of the interval ([-3, -4]), which is 1.
Hence, ∫[-3, -4] -4 dt = -4 * 1 = -4.
7. g(-3) = -4.
8. Determine g(0):
Since 0 < 4, g(0) falls under the first part of f(x) where f(x) = 0.
Thus, g(0) = ∫[0, -4] f(t)dt = ∫[0, -4] 0 dt.
9. Evaluate ∫[0, -4] 0 dt:
Here, the integrand is 0, and integrating it over any interval will always result in 0.
Therefore, ∫[0, -4] 0 dt = 0.
10. g(0) = 0.
11. Calculate g(6):
Since 5 ≤ 6, g(6) is covered by the fourth part of f(x) where f(x) = 0.
Hence, g(6) = ∫[6, -4] f(t)dt = ∫[6, -4] 0 dt.
12. Evaluate ∫[6, -4] 0 dt:
Again, the integrand is 0, resulting in 0 regardless of the interval chosen.
Therefore, ∫[6, -4] 0 dt = 0.
13. g(6) = 0.
By analyzing the function f(x) and computing g(x) using the given steps, we have determined the values for g(-5) = 5, g(-3) = -4, g(0) = 0, and g(6) = 0.
As for the absolute maximum of g(x), we can observe that the function g(x) is constant and equal to 5 whenever x < -4 or -4 ≤ x < -1. Therefore, the absolute maximum of g(x) occurs when x < -4 or -4 ≤ x < -1, and the value is 5.
To determine the values of g(-5), g(-3), g(0), and g(6), we need to evaluate the definite integral of f(t) from the lower limit of -4 to each corresponding x value.
Let's start by finding g(-5). The lower limit of integration is -4, and the upper limit is -5. Since -5 is less than -4, g(-5) will be equal to the integral of f(t) from -4 to -5.
For x < -4, f(x) = 0. Therefore, the integral of f(t) from -4 to -5 will be equal to 0.
Next, let's find g(-3). Again, the lower limit of integration is -4, and the upper limit is -3. Since -3 is within the range where f(t) has a defined value, in this case, -4, g(-3) will be equal to the integral of f(t) from -4 to -3.
For -4 ≤ x < -1, f(x) = 5. Therefore, the integral of f(t) from -4 to -3 will be equal to 5.
Moving on to g(0), we have a lower limit of -4 and an upper limit of 0. Since 0 is within the range where f(t) has a defined value, in this case, -4, g(0) will be equal to the integral of f(t) from -4 to 0.
For -4 ≤ x < 5, f(x) = -4. Therefore, the integral of f(t) from -4 to 0 will be equal to -4 times the width of the interval, which is 4. So, g(0) = -4 * 4 = -16.
Lastly, let's calculate g(6), with a lower limit of -4 and an upper limit of 6. Since 6 is greater than or equal to 5, g(6) will be equal to the integral of f(t) from -4 to 5.
For x ≥ 5, f(x) = 0. Therefore, the integral of f(t) from -4 to 5 will be equal to 0.
Now, to find the absolute maximum of g(x), we need to evaluate g(x) for all possible x values. So, we calculate g(-5), g(-3), g(0), and g(6).
We obtained the following values:
g(-5) = 0
g(-3) = 5
g(0) = -16
g(6) = 0
The absolute maximum of g(x) occurs when x = -3, and the value at x = -3 is 5.