Example Question - area under the curve
Here are examples of questions we've helped users solve.
Estimating Distance Traveled by a Train
The image shows a graph of the velocity of a train against time. To solve questions a) and b), you will need to estimate the area under the velocity-time curve for the first 20 seconds, as the area under a velocity-time graph represents displacement (distance traveled) for a given time interval. a) To use four strips of equal width to estimate the distance the train traveled in the first 20 seconds, divide the time interval (0-20 seconds) into four equal parts. Each part is 5 seconds long. For each 5-second interval, we calculate the area of the rectangular strip, which will be approximately the velocity at the start of the interval multiplied by the time duration (width of the strip). 1. For the first interval (0-5 seconds), the velocity looks to be about 4 m/s. Area = velocity × time = 4 m/s × 5 s = 20 m 2. For the second interval (5-10 seconds), the velocity seems to be roughly 10 m/s. Area = 10 m/s × 5 s = 50 m 3. For the third interval (10-15 seconds), the velocity is around 18 m/s. Area = 18 m/s × 5 s = 90 m 4. For the fourth interval (15-20 seconds), the velocity is about 24 m/s. Area = 24 m/s × 5 s = 120 m Add the areas for each interval to get the total estimated displacement: Total displacement ≈ 20 m + 50 m + 90 m + 120 m = 280 m This is an estimation, and the actual distance might be slightly different. b) The graph shows a curve that indicates increasing acceleration, meaning the velocity of the train is not constant but increasing as time passes. Our rectangular strips underestimated the velocity (and thus distance) for the latter part of each time interval. Given that the velocity of the train is increasing, the actual area (and therefore the actual distance traveled) will be greater than what the rectangles estimate because the velocity at the end of each interval is higher than at the beginning. Thus, our rectangular estimation using the velocities at the start of each interval is an underestimate.
Probability Calculation Using Probability Density Function
The image shows a graph with a density function \( f(x) \) for a random variable X and asks to find the probability that X is between 1 and 4, \( P(1 < X < 4) \). The graph depicts a triangle that represents the probability density function, with the base of the triangle spanning from x = 0 to x = 4 and the height at f(x) = 1/2 at x = 0. The area under the curve of a probability density function \( f(x) \) between two points a and b gives the probability \( P(a < X < b) \). To find \( P(1 < X < 4) \), we will calculate the area of the triangle between x = 1 and x = 4. The original triangle's base is 4 units long (from x = 0 to x = 4), and the height is \( f(x) = 1/2 \). The area of the full triangle is: \( \text{Area} = \frac{1}{2} (\text{base}) (\text{height}) = \frac{1}{2} (4) \left(\frac{1}{2}\right) = 1 \) Now, we need only the area from x = 1 to x = 4. To find this portion, we subtract the area of the smaller triangle from x = 0 to x = 1 from the total area. The smaller triangle has a base of 1 unit and the same height of \( f(x) = 1/2 \). The area of this smaller triangle is: \( \text{Area}_{\text{small}} = \frac{1}{2} (\text{base of small triangle}) (\text{height}) = \frac{1}{2} (1) \left(\frac{1}{2}\right) = \frac{1}{4} \) Now subtract the area of the small triangle from the total area to find the desired probability: \( P(1 < X < 4) = \text{Area}_{\text{total}} - \text{Area}_{\text{small}} = 1 - \frac{1}{4} = \frac{3}{4} \) So, the probability that X is between 1 and 4 is \( \frac{3}{4} \) or 0.75.
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