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Ch. 5 - Integration
5์žฅ, ๋ฌธ์ œ 5.3.85

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 ) .

๊ฒ€์ฆ๋œ ๋‹จ๊ณ„๋ณ„ ์•ˆ๋‚ด
1
Step 1: Recognize that the problem involves the Fundamental Theorem of Calculus, which states that if F(x) = โˆซโ‚หฃ f(t) dt, then dF/dx = f(x). This theorem will be key in solving the derivative of the integral.
Step 2: Analyze the given integral โˆซโ‚€หฃ (โˆš1 + tยฒ) dt. According to the Fundamental Theorem of Calculus, the derivative of this integral with respect to x is simply the integrand evaluated at the upper limit of integration, which is โˆš(1 + xยฒ).
Step 3: Consider the hint provided: โˆซหฃโ‚‹โ‚“ (โˆš1 + tยฒ) dt = โˆซโฐโ‚‹โ‚“ (โˆš1 + tยฒ) dt + โˆซหฃโ‚‹โ‚“ (โˆš1 + tยฒ) dt. This suggests breaking the integral into parts, but for the derivative d/d๐“ โˆซโ‚€หฃ (โˆš1 + tยฒ) dt, the hint is not directly necessary since the Fundamental Theorem of Calculus simplifies the process.
Step 4: Apply the Fundamental Theorem of Calculus directly to the integral โˆซโ‚€หฃ (โˆš1 + tยฒ) dt. The derivative with respect to x is simply โˆš(1 + xยฒ), as the lower limit of integration (0) does not contribute to the derivative.
Step 5: Conclude that the derivative of the given integral is โˆš(1 + xยฒ). The hint provided is more relevant for breaking down integrals with different limits, but in this case, the direct application of the theorem suffices.

๋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:
1m
๋„์›€์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 differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then the integral of f from a to b can be computed as F(b) - F(a). This theorem also implies that the derivative of an integral function is the integrand evaluated at the upper limit of integration.
์ถ”์ฒœ ์˜์ƒ:
๊ฐ€์ด๋“œ ์ฝ”์Šค
06:11
Fundamental Theorem of Calculus Part 1

Differentiation under the Integral Sign

Differentiation under the integral sign allows us to differentiate an integral with respect to a parameter. This technique is useful when the limits of integration or the integrand itself depend on a variable, enabling the evaluation of complex integrals by treating them as functions of that variable.
์ถ”์ฒœ ์˜์ƒ:

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is expressed as โˆซu dv = uv - โˆซv du. This method can simplify the integration of more complex expressions, particularly when one function is easily integrable and the other is easily differentiable.
์ถ”์ฒœ ์˜์ƒ:
๊ฐ€์ด๋“œ ์ฝ”์Šค
06:18
Integration by Parts for Definite Integrals
๊ด€๋ จ ์‹ค์ฒœ
๊ต๊ณผ์„œ ์งˆ๋ฌธ

Average value of the derivative Suppose ฦ’ ' is a continuous function for all real numbers. Show that the average value of the derivative on an interval [a, b] is ฦ’โป' = (ฦ’(b) โ€•ฦ’(a))/ (bโ€•a) . Interpret this result in terms of secant lines.

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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

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. 

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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

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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