Textbook Question
Definite integrals from graphs The figure shows the areas of regions bounded by the graph of ƒ and the 𝓍-axis. Evaluate the following integrals.
∫ₐᶜ ƒ(𝓍) d𝓍
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9. Graphical Applications of Integrals
Area Between Curves
4:22 minutesProblem 8.4.83
Textbook Question
Visual proof Let F(x)=∫₀ˣ √(a²−t²) dt. The figure shows that F(x)= area of sector OAB+ area of triangle OBC.
a. Use the figure to prove that
F(x) = (a² sin ⁻¹(x/a))/2 + x√(a²−x²)/2
b. Conclude that ∫ √(a²−x²) dx = (a² sin ⁻¹(x/a))/2 + x√(a²−x²)/2 + C.
Verified step by step guidance1
Step 1: Recognize that the function \(F(x) = \int_0^x \sqrt{a^2 - t^2} \, dt\) represents the area under the curve \(y = \sqrt{a^2 - t^2}\) from \(t=0\) to \(t=x\). The figure shows this area as the sum of two parts: the area of sector \(OAB\) and the area of triangle \(OBC\).
Step 2: Identify the angle \(\theta\) in the figure, where \(\theta = \sin^{-1}(x/a)\). This comes from the right triangle \(OBC\) where \(OC = x\) and \(OB = a\), so \(\sin \theta = \frac{x}{a}\).
Step 3: Calculate the area of sector \(OAB\). Since \(OAB\) is a sector of a circle with radius \(a\) and central angle \(\theta\), its area is given by \(\frac{1}{2} a^2 \theta = \frac{1}{2} a^2 \sin^{-1}(x/a)\).
Step 4: Calculate the area of triangle \(OBC\). The base \(OC\) is \(x\) and the height \(BC\) is \(\sqrt{a^2 - x^2}\) (from the Pythagorean theorem). Therefore, the area of triangle \(OBC\) is \(\frac{1}{2} \times x \times \sqrt{a^2 - x^2}\).
Step 5: Add the two areas to express \(F(x)\) as the sum of the sector and triangle areas: \(F(x) = \frac{1}{2} a^2 \sin^{-1}(x/a) + \frac{1}{2} x \sqrt{a^2 - x^2}\). This completes the proof for part (a). For part (b), since \(F'(x) = \sqrt{a^2 - x^2}\), the integral \(\int \sqrt{a^2 - x^2} \, dx\) equals \(F(x) + C\), giving the formula \(\int \sqrt{a^2 - x^2} \, dx = \frac{1}{2} a^2 \sin^{-1}(x/a) + \frac{1}{2} x \sqrt{a^2 - x^2} + C\).
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Video duration:
4mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integral as Area Under a Curve
The definite integral of a function from 0 to x represents the area under the curve between these limits. In this problem, F(x) = ∫₀ˣ √(a²−t²) dt corresponds to the area under the semicircle y = √(a²−t²), which can be geometrically interpreted as the sum of areas of simpler shapes.
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Area of a Circular Sector
A sector of a circle is a portion bounded by two radii and the arc between them. Its area is given by (1/2)r²θ, where r is the radius and θ is the central angle in radians. Here, the sector OAB with radius a and angle θ = sin⁻¹(x/a) helps express part of the integral's area.
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Area of a Triangle Using Trigonometry
The area of a triangle can be found using (1/2) base × height or (1/2)ab sin C for two sides a, b and included angle C. In this problem, triangle OBC's area is computed using the base x and height √(a²−x²), which complements the sector area to represent the integral.
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