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Ch. 5 - Integration
5์žฅ, ๋ฌธ์ œ 5.R.113c

Function defined by an integral Let ฦ’(๐“) = โˆซโ‚€หฃ (t โ€• 1)ยนโต (tโ€•2)โน dt .
(c) For what values of ๐“ does ฦ’ have local minima? Local maxima?

๊ฒ€์ฆ๋œ ๋‹จ๊ณ„๋ณ„ ์•ˆ๋‚ด
1
Step 1: Recall that to find local minima and maxima of a function, we need to analyze its derivative. The Fundamental Theorem of Calculus tells us that the derivative of ฦ’(๐“) = โˆซโ‚€หฃ g(t) dt is ฦ’'(๐“) = g(๐“). Here, g(t) = (t - 1)ยนโต (t - 2)โน.
Step 2: Set ฦ’'(๐“) = g(๐“) = (๐“ - 1)ยนโต (๐“ - 2)โน equal to zero to find critical points. This equation is satisfied when either (๐“ - 1) = 0 or (๐“ - 2) = 0. Thus, the critical points are ๐“ = 1 and ๐“ = 2.
Step 3: To determine whether these critical points correspond to local minima or maxima, analyze the sign changes of ฦ’'(๐“) = (๐“ - 1)ยนโต (๐“ - 2)โน around the critical points. Consider intervals around ๐“ = 1 and ๐“ = 2, such as (0, 1), (1, 2), and (2, โˆž).
Step 4: Evaluate the behavior of ฦ’'(๐“) in each interval. For example, in the interval (0, 1), both (๐“ - 1)ยนโต and (๐“ - 2)โน are negative, making ฦ’'(๐“) positive. In the interval (1, 2), (๐“ - 1)ยนโต is positive and (๐“ - 2)โน is negative, making ฦ’'(๐“) negative. In the interval (2, โˆž), both terms are positive, making ฦ’'(๐“) positive.
Step 5: Based on the sign changes of ฦ’'(๐“), conclude that ๐“ = 1 is a local maximum (since ฦ’'(๐“) changes from positive to negative) and ๐“ = 2 is a local minimum (since ฦ’'(๐“) changes from negative to positive).

๋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:
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๋„์›€์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.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation and integration, stating that if a function is defined as an integral, its derivative can be found by evaluating the integrand at the upper limit of integration. This theorem is essential for analyzing the behavior of the function ฦ’(๐“) in the given question.
์ถ”์ฒœ ์˜์ƒ:
๊ฐ€์ด๋“œ ์ฝ”์Šค
06:11
Fundamental Theorem of Calculus Part 1

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are crucial for determining local maxima and minima, as they represent potential locations where the function's behavior changes. In the context of ฦ’(๐“), finding where the derivative equals zero will help identify these critical points.
์ถ”์ฒœ ์˜์ƒ:
04:50
Critical Points

Second Derivative Test

The Second Derivative Test is a method used to classify critical points as local minima, local maxima, or saddle points. By evaluating the second derivative at a critical point, one can determine the concavity of the function at that point. If the second derivative is positive, the point is a local minimum; if negative, it is a local maximum.
์ถ”์ฒœ ์˜์ƒ:
06:02
The Second Derivative Test: Finding Local Extrema
๊ด€๋ จ ์‹ค์ฒœ
๊ต๊ณผ์„œ ์งˆ๋ฌธ

(b) Find the average value of ฦ’ shown in the figure on the interval [2,6] and then find the point(s) c in (2, 6) guaranteed to exist by the Mean Value Theorem for Integrals. 

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Geometry of integrals Without evaluating the integrals, explain why the following statement is true for positive integers n:

โˆซโ‚€ยน ๐“โฟd๐“ + โˆซโ‚€ยน โฟโˆš(๐“d๐“) = 1

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Area of regions Compute the area of the region bounded by the graph of ฦ’ and the ๐“-axis on the given interval. You may find it useful to sketch the region.                                              

                                                                                                                                                                                    

 ฦ’(๐“) = 2 sin ๐“/4 on [0, 2ฯ€]

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Symmetry properties Suppose โˆซโ‚€โด ฦ’(๐“) d๐“ = 10 and โˆซโ‚€โด g(๐“) d๐“ = 20. Furthermore, suppose ฦ’ is an even function and g is an odd function. Evaluate the following integrals.


(e) โˆซโ‚‹โ‚‚ยฒ 3๐“ฦ’(๐“)d๐“

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Symmetry properties Suppose โˆซโ‚€โด ฦ’(๐“) d๐“ = 10 and โˆซโ‚€โด g(๐“) d๐“ = 20. Furthermore, suppose ฦ’ is an even function and g is an odd function. Evaluate the following integrals.


(a) โˆซโ‚‹โ‚„โด ฦ’(๐“) d๐“

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Symmetry properties Suppose โˆซโ‚€โด ฦ’(๐“) d๐“ = 10 and โˆซโ‚€โด g(๐“) d๐“ = 20. Furthermore, suppose ฦ’ is an even function and g is an odd function. Evaluate the following integrals.


(c) โˆซโ‚‹โ‚„โด (4ฦ’(๐“) โ€• 3g(๐“))d๐“

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