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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.3.85
Chapter 5, Problem 5.3.85

Derivatives of integrals Simplify the following expressions.


d/d𝓍 βˆ«β‚€Λ£ (√1 + tΒ²) dt (Hint: βˆ«Λ£β‚‹β‚“ (√1 + tΒ²) dt = βˆ«β°β‚‹β‚“ (√1 + tΒ²) dt + βˆ«Λ£β‚‹β‚“ (√1 + tΒ²) dt ) .

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Step 1: Recognize that the problem involves the Fundamental Theorem of Calculus, which states that if F(x) = βˆ«β‚Λ£ 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 βˆ«β‚€Λ£ (√1 + tΒ²) 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 √(1 + xΒ²).
Step 3: Consider the hint provided: βˆ«Λ£β‚‹β‚“ (√1 + tΒ²) dt = βˆ«β°β‚‹β‚“ (√1 + tΒ²) dt + βˆ«Λ£β‚‹β‚“ (√1 + tΒ²) dt. This suggests breaking the integral into parts, but for the derivative d/d𝓍 βˆ«β‚€Λ£ (√1 + tΒ²) 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 βˆ«β‚€Λ£ (√1 + tΒ²) dt. The derivative with respect to x is simply √(1 + xΒ²), as the lower limit of integration (0) does not contribute to the derivative.
Step 5: Conclude that the derivative of the given integral is √(1 + x²). The hint provided is more relevant for breaking down integrals with different limits, but in this case, the direct application of the theorem suffices.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.
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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.
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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 ∫u dv = uv - ∫v du. This method can simplify the integration of more complex expressions, particularly when one function is easily integrable and the other is easily differentiable.
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Related Practice
Textbook Question

Derivatives of integrals Simplify the following expressions.


d/dt βˆ«β‚€α΅— d𝓍/(1 + 𝓍²) + βˆ«β‚ΒΉ/α΅— dx/(1 + 𝓍²)

Textbook Question

Definite integrals from graphs The figure shows the areas of regions bounded by the graph of Ζ’ and the 𝓍-axis. Evaluate the following integrals.



βˆ«β‚€α΅ƒ Ζ’(𝓍) d𝓍

Textbook Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 βˆ« (sin⁡ 𝓍 + 3 sinΒ³ 𝓍― sin 𝓍) cos 𝓍 d𝓍

Textbook Question

Average distance on a triangle Consider the right triangle with vertices (0,0) ,(0,b) , and (a,0) , where a > 0 and b > 0. Show that the average vertical distance from points on the 𝓍-axis to the hypotenuse is b/2 , for all a > 0 .

Textbook Question

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.

βˆ«α΅ƒβ‚‹β‚ Ζ’(p(𝓍)) d𝓍

Textbook Question

Left and right Riemann sums Use the figures to calculate the left and right Riemann sums for f on the given interval and for the given value of n.

Ζ’(𝓍) = x + 1 on [1,6] ; n = 5

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