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Ch. 5 - Integration
5์žฅ, ๋ฌธ์ œ 5.3.5

The linear function ฦ’(๐“) = 3 โ€• ๐“ is decreasing on the interval [0, 3]. Is its area function for ฦ’ (with left endpoint 0) increasing or decreasing on the interval [0, 3]? Draw a picture and explain. 

๊ฒ€์ฆ๋œ ๋‹จ๊ณ„๋ณ„ ์•ˆ๋‚ด
1
Step 1: Understand the problem. The linear function ฦ’(๐“) = 3 - ๐“ is given, and we are tasked to determine whether its area function (integral) is increasing or decreasing on the interval [0, 3]. The area function represents the accumulated area under the curve of ฦ’(๐“) starting from the left endpoint 0.
Step 2: Recall the relationship between a function and its area function. The area function is the integral of ฦ’(๐“) with respect to ๐“. Mathematically, the area function A(๐“) is defined as: Ax=0f(x)dx. The derivative of the area function, A'(๐“), is equal to ฦ’(๐“).
Step 3: Analyze the behavior of ฦ’(๐“) on the interval [0, 3]. The function ฦ’(๐“) = 3 - ๐“ is linear with a negative slope (-1), meaning it decreases as ๐“ increases. Specifically, ฦ’(๐“) starts at 3 when ๐“ = 0 and decreases to 0 when ๐“ = 3.
Step 4: Determine the behavior of the area function A(๐“). Since A'(๐“) = ฦ’(๐“), the area function A(๐“) increases wherever ฦ’(๐“) is positive. On the interval [0, 3], ฦ’(๐“) is positive (above the x-axis), so the area function A(๐“) is increasing throughout this interval.
Step 5: Visualize the problem. Draw the graph of ฦ’(๐“) = 3 - ๐“, which is a straight line decreasing from (0, 3) to (3, 0). Shade the area under the curve from ๐“ = 0 to a variable endpoint ๐“. As ๐“ moves from 0 to 3, the shaded area grows, confirming that the area function A(๐“) is increasing on [0, 3].

๋น„์Šทํ•œ ๋ฌธ์ œ์— ๋Œ€ํ•œ ๊ฒ€์ฆ๋œ ์˜์ƒ ๋‹ต๋ณ€:

์ด ์˜์ƒ ํ•ด๋ฒ•์€ ์œ„ ๋ฌธ์ œ์— ๋„์›€์ด ๋œ๋‹ค๊ณ  ํŠœํ„ฐ๋“ค์ด ์ถ”์ฒœํ•œ ๊ฒƒ์ž…๋‹ˆ๋‹ค.
์˜์ƒ ๊ธธ์ด:
3m
๋„์›€์ด ๋˜์—ˆ๋‚˜์š”?

์ฃผ์š” ๊ฐœ๋…

์งˆ๋ฌธ์— ์˜ฌ๋ฐ”๋ฅด๊ฒŒ ๋‹ตํ•˜๊ธฐ ์œ„ํ•ด ๋ฐ˜๋“œ์‹œ ์ดํ•ดํ•ด์•ผ ํ•˜๋Š” ํ•ต์‹ฌ ๊ฐœ๋…๋“ค์€ ๋‹ค์Œ๊ณผ ๊ฐ™์Šต๋‹ˆ๋‹ค.

Decreasing Functions

A function is considered decreasing on an interval if, for any two points x1 and x2 within that interval, where x1 < x2, the function value at x1 is greater than the function value at x2 (ฦ’(x1) > ฦ’(x2)). In this case, the linear function ฦ’(๐“) = 3 - ๐“ decreases as x increases, indicating that as we move from 0 to 3, the output values of the function get smaller.
์ถ”์ฒœ ์˜์ƒ:
07:32
Determining Where a Function is Increasing & Decreasing

Area Function

The area function A(x) associated with a function ฦ’(๐“) represents the accumulated area under the curve of ฦ’ from a starting point (in this case, 0) to a variable endpoint x. Mathematically, it is defined as A(x) = โˆซ[0,x] ฦ’(t) dt. The behavior of the area function depends on the values of the original function; if ฦ’ is decreasing, the area function will reflect this change.
์ถ”์ฒœ ์˜์ƒ:
05:06
Finding Area When Bounds Are Not Given

Relationship Between Function and Area Function

The relationship between a function and its area function is governed by the Fundamental Theorem of Calculus. If the original function is decreasing, the area function will increase at a decreasing rate. This means that while the area function A(x) is increasing as x moves from 0 to 3, the rate of increase diminishes because the heights of the rectangles (representing area) are getting smaller as the function value decreases.
์ถ”์ฒœ ์˜์ƒ:
05:23
Finding Area Between Curves on a Given Interval
๊ด€๋ จ ์‹ค์ฒœ
๊ต๊ณผ์„œ ์งˆ๋ฌธ

General results Evaluate the following integrals in which the function ฦ’ is unspecified. Note that ฦ’โฝแต–โพ is the pth derivative of ฦ’ and ฦ’แต– is the pth power of ฦ’. Assume ฦ’ and its derivatives are continuous for all real numbers. 

โˆซ (5 ฦ’ยณ (๐“) + 7ฦ’ยฒ (๐“) + ฦ’ (๐“ )) ฦ’'(๐“) d๐“

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

{Use of Tech} Areas of regions Find the area of the region ๐‘… bounded by the graph of ฦ’ and the ๐“-axis on the given interval. Graph ฦ’ and show the region ๐‘….                                              

                                                                                                                                                                                    

 ฦ’(๐“) = 2 โ€• |๐“| on [ โ€• 2 , 4]

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Why can the constant of integration be omitted from the antiderivative when evaluating a definite integral?

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Approximating displacement The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into n subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles.

v = 2t + 1(m/s), for 0 โ‰ค t โ‰ค 8 ; n = 2

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Derivatives of integrals Simplify the following expressions.


d/d๐“ โˆซโ‚€หฃ (โˆš1 + tยฒ) dt (Hint: โˆซหฃโ‚‹โ‚“ (โˆš1 + tยฒ) dt = โˆซโฐโ‚‹โ‚“ (โˆš1 + tยฒ) dt + โˆซหฃโ‚‹โ‚“ (โˆš1 + tยฒ) dt ) .

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Symmetry of composite functions Prove that the integrand is either even or odd. Then give the value of the integral or show how it can be simplified. Assume f and g are even functions and p and q are odd functions.

โˆซแตƒโ‚‹โ‚ ฦ’(p(๐“)) d๐“

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