Textbook Question
Areas of regions Find the area of the region bounded by the graph of Ζ and the π-axis on the given interval.
Ζ(π) = πΒ³ β 1 on [β1, 2]
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9. Graphical Applications of Integrals
Area Between Curves
3:46 minutesProblem 5.R.89
Textbook Question
Area of regions Compute the area of the region bounded by the graph of Ζ and the π-axis on the given interval. You may find it useful to sketch the region.
Ζ(π) = 2 sin π/4 on [0, 2Ο]
Verified step by step guidance1
Step 1: Understand the problem. We are tasked with finding the area of the region bounded by the graph of Ζ(π) = 2 sin(π/4) and the π-axis over the interval [0, 2Ο]. This involves integrating the function over the given interval.
Step 2: Set up the integral. The area can be computed using the definite integral of the absolute value of Ζ(π) over the interval [0, 2Ο]. Since the sine function can be negative, we need to account for this by splitting the integral into regions where Ζ(π) is positive and negative. The integral is: β«[0, 2Ο] |2 sin(π/4)| dπ.
Step 3: Determine where the function changes sign. To find where Ζ(π) = 2 sin(π/4) changes sign, solve sin(π/4) = 0. The solutions occur at π/4 = nΟ, where n is an integer. Over the interval [0, 2Ο], this corresponds to π = 0, Ο, and 2Ο. These points divide the interval into subintervals: [0, Ο] and [Ο, 2Ο].
Step 4: Evaluate the integral over each subinterval. On [0, Ο], sin(π/4) is positive, so the integral is β«[0, Ο] 2 sin(π/4) dπ. On [Ο, 2Ο], sin(π/4) is negative, so the integral is β«[Ο, 2Ο] -2 sin(π/4) dπ. Combine these results to compute the total area.
Step 5: Solve each integral. Use the substitution u = π/4, which implies du = 1/4 dπ. Rewrite the integrals in terms of u, evaluate them, and sum the results to find the total area. Remember to substitute back to the original variable after integration.
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Video duration:
3mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integral
The definite integral of a function over a specified interval represents the net area between the graph of the function and the x-axis. It is calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. For the function f(x) = 2 sin(x/4), the definite integral from 0 to 2Ο will yield the total area of the region bounded by the curve and the x-axis within that interval.
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Area Under the Curve
The area under the curve of a function can be positive or negative depending on whether the function is above or below the x-axis. When calculating the area, it is essential to consider the absolute value of the integral if the function dips below the x-axis, as this affects the total area calculation. In this case, the area can be found by integrating the function and taking the absolute value of any negative contributions.
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Graphical Interpretation
Sketching the graph of the function provides a visual representation of the area to be calculated. For f(x) = 2 sin(x/4), plotting the function over the interval [0, 2Ο] helps identify where the function intersects the x-axis, which is crucial for determining the limits of integration and understanding the shape of the area. This graphical approach aids in visualizing the contributions to the area from different segments of the curve.
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