9. Graphical Applications of Integrals

Area Between Curves

9. Graphical Applications of Integrals

Area Between Curves

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Multiple Choice

Calculate the area of the shaded region between the 2 functions from $x equals 0$ to $x equals 9$

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Step 1: Identify the two functions that bound the shaded region. From the graph, the upper function is f(x) = √x and the lower function is g(x) = x/3.

Step 2: Set up the integral to calculate the area between the two curves. The area is given by the integral of the difference between the upper function and the lower function over the interval [0, 9]. This can be expressed as: ∫[0,9] (√x - x/3) dx.

Step 3: Break down the integral into two separate integrals for easier computation: ∫[0,9] √x dx - ∫[0,9] (x/3) dx.

Step 4: Compute the antiderivative of each term. For √x, rewrite it as x^(1/2) and use the power rule to find the antiderivative: (2/3)x^(3/2). For x/3, factor out 1/3 and use the power rule to find the antiderivative: (1/6)x^2.

Step 5: Evaluate the definite integrals by substituting the limits of integration (x = 0 and x = 9) into the antiderivatives. Subtract the results of the two integrals to find the total area of the shaded region.

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Textbook Question

Definitions of hyperbolic sine and cosine Complete the following steps to prove that when the x- and y-coordinates of a point on the hyperbola x² - y² = 1 are defined as cosh t and sinh t, respectively, where t is twice the area of the shaded region in the figure, x and y can be expressed as

x = cosh t = (eᵗ + e⁻ᵗ) / 2 and y = sinh t = (eᵗ - e⁻ᵗ) / 2.

b. In Chapter 8, the formula for the integral in part (a) is derived:

∫ √(z² − 1) dz = (z/2)√(z² − 1) − (1/2) ln|z + √(z² − 1)| + C.

Evaluate this integral on the interval [1, x], explain why the absolute value can be dropped, and combine the result with part (a) to show that:

t = ln(x + √(x² − 1)).

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Textbook Question

Area of a sector of a hyperbola: Consider the region R bounded by the right branch of the hyperbola x²/a² - y²/b² = 1 and the vertical line through the right focus

a. What is the area of R?

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Textbook Question

Probability as an integral Two points P and Q are chosen randomly, one on each of two adjacent sides of a unit square (see figure). What is the probability that the area of the triangle formed by the sides of the square and the line segment PQ is less than one-fourth the area of the square? Begin by showing that x and y must satisfy xy < 1/2 in order for the area condition to be met. Then argue that the required probability is: 1/2 + ∫[1/2 to 1] (dx / 2x) and evaluate the integral.

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Textbook Question

A power line is attached at the same height to two utility poles that are separated by a distance of 100 ft; the power line follows the curve ƒ(x) = a cosh x/a. Use the following steps to find the value of a that produces a sag of 10 ft midway between the poles. Use a coordinate system that places the poles at x = ±50.

a. Show that a satisfies the equation cosh 50/a − 1 = 10/a.

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Multiple Choice

Calculate the area of the shaded region between $f (x)$ & $g (x)$ contained between $x = - 4$ & $x = - 2$ .

Sketch the region bounded by $f (x) = - {(x - 2)}^{2} + 5$ & $g of x equals 4 x$ on the interval $open bracket 0 comma 1 close bracket$ . Then set up an integral to represent the region's area and evaluate.

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Multiple Choice

Shade the region bounded by $f (x) = \sin x$ & $g of x equals cosine 2 x$ on the interval $[\frac{π}{6}, \frac{5 π}{6}]$ . Then set up an integral to represent the region's area.