Definite integral of a function f(x) between two limits x=a and x=b denotesReview LaterArea bounded by x=a, x-axis, f(x) and x=bArea bounded by x=b, y-axis, x-axis and f(x)Area bounded by x=a, y-axis, x-axis and f(x)Area bounded by x=f(a), x=f(b), x-axis and f(x)
Question
Definite integral of a function f(x) between two limits x=a and x=b denotesReview LaterArea bounded by x=a, x-axis, f(x) and x=bArea bounded by x=b, y-axis, x-axis and f(x)Area bounded by x=a, y-axis, x-axis and f(x)Area bounded by x=f(a), x=f(b), x-axis and f(x)
Solution
The definite integral of a function f(x) between two limits x=a and x=b denotes the area bounded by x=a, x-axis, f(x) and x=b. This is a fundamental concept in calculus, specifically integral calculus. Here's a step-by-step explanation:
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The definite integral is a mathematical concept that is used to calculate the area under a curve, between two points on the x-axis (a and b).
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The function f(x) represents the curve in the xy-plane. The x-axis and the lines x=a and x=b form the boundaries of the area.
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The area is calculated by integrating the function f(x) from a to b. This is denoted as ∫ from a to b of f(x) dx.
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The result of this integral gives the exact area under the curve f(x) from x=a to x=b.
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If the function f(x) is above the x-axis for the interval [a, b], the definite integral is positive. If f(x) is below the x-axis, the definite integral is negative.
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If the function crosses the x-axis, the definite integral calculates the net area, which is the difference between the area above the x-axis and the area below the x-axis.
So, the definite integral of a function f(x) between two limits x=a and x=b denotes the net area bounded by x=a, x-axis, f(x) and x=b.
Similar Questions
The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.(a) 10f(x) dx0 (b) 25f(x) dx0 (c) 35f(x) dx25 (d) 45f(x) dx0
What is the difference between definite and indefinite integrals?
EXAMPLE 4 Evaluate the following integrals by interpreting each in terms of areas.(a) 39 − x2 dx0(b) 10(x − 3) dx0SOLUTION (a) Since f(x) = 9 − x2 ≥ 0, we can interpret this integral as the area under the curve y = 9 − x2 from 0 to . But since y2 = , we get x2 + y2 = 9, which shows that the graph of f is a quarter-circle with radius in the top figure. Therefore,39 − x2 dx0 = 14𝜋(3)2 = .(b) The graph of y = x − 3 is the line with slope shown in the bottom figure. We compute the integral as the difference of the areas of the two triangles:10(x − 3) dx0 = A1 − A2 = − 4.5 = .
he integral represents the area between the two curves f(x)=x+4𝑓(𝑥)=𝑥+4 and g(x)=4−x2𝑔(𝑥)=4−𝑥2 and between left and right boundaries for x as shown in the Figure below.Geogebra plot. You can zoom with the mouse wheel and move the graph by clicking on the background and dragging.__________________________________________________a) Express the area as a definite integral:
Evaluate the definite integral. Use a graphing utility to verify your result.54x3x − 4 dx
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