8.2first Fundamental Theorem Of Calculus Ap Calculus 2
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Fundamental Theorem of Calculus Student Session-Presenter Notes This session includes a reference sheet at the back of the packet. We suggest that the presenter not spend time going over the reference sheet, but point it out to students so that they may refer to it if needed. 4.4 The Fundamental Theorem of Calculus Section Project: Demonstrating the Fundamental Theorem 4.5 Integration by Substitution 4.6 Numerical Integration. Logarithmic, Exponential, and Other Transcendental Functions. 5.1 The Natural Logarithmic Function: Differentiation 5.2 The Natural Logarithmic Function: Integration 5.3 Inverse Functions.

Following are some of the most frequently used theorems, formulas, and definitions that you encounter in a calculus class for a single variable. The list isn’t comprehensive, but it should cover the items you’ll use most often.

Limit Definition of a Derivative

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Definition: Continuous at a number a

The Intermediate Value Theorem

Definition of a Critical Number

A critical number of a function f is a number c in the domain of f such that either f‘(c)= 0 or f‘(c) does not exist.

Rolle’s Theorem

Let f be a function that satisfies the following three hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).
f(a)= f(b).

Then there is a number c in (a, b) such that f‘(c)= 0.

The Mean Value Theorem

Let f be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Newton’s Method Approximation Formula

Newton’s method is a technique that tries to find a root of an equation. To begin, you try to pick a number that’s “close” to the value of a root and call this value x₁. Picking x₁ may involve some trial and error; if you’re dealing with a continuous function on some interval (or possibly the entire real line), the intermediate value theorem may narrow down the interval under consideration. After picking x₁, you use the recursive formula given here to find successive approximations:

A word of caution: Always verify that your final approximation is correct (or close to the value of the root). Newton’s method can fail in some instances, based on the value picked for x₁. Any calculus text that covers Newton’s method should point out these shortcomings.

The Fundamental Theorem of Calculus

Suppose f is continuous on [a, b]. Then the following statements are true:

The Trapezoid Rule

where

Simpson’s Rule

where n is even and

The Fundamental Theorem of Calculus (FTC) shows that differentiation and integration are inverse processes.

Part (1) (FTC1)

If (f) is a continuous function on (left[ {a,b} right],) then the function (g) defined by

[{{gleft( x right)} = intlimits_a^x {fleft( {t} right)dt},;;}kern0pt{{a le x le b}}]

is an antiderivative of (f), that is

[{g^primeleft( x right) = fleft( x right);;text{or};;}kern0pt{frac{d}{{dx}}left( {intlimits_a^x {fleft( t right)dt} } right) }={ fleft( x right).}]

If (f) happens to be a positive function, then (gleft( x right)) can be interpreted as the area under the graph of (f) from (a) to (x.)

The first part of the theorem says that if we first integrate (f) and then differentiate the result, we get back to the original function (f.)

Part (2) (FTC2)

The second part of the fundamental theorem tells us how we can calculate a definite integral.

If (f) is a continuous function on (left[ {a,b} right]) and (F) is an antiderivative of (f,) that is (F^prime = f,) then

[{intlimits_a^b {fleft( x right)dx} }= {Fleft( b right) – Fleft( a right);;}kern0pt{text{or};;{intlimits_a^b {{F^primeleft( x right)}dx} }= {Fleft( b right) – Fleft( a right)}.}]

To evaluate the definite integral of a function (f) from (a) to (b,) we just need to find its antiderivative (F) and compute the difference between the values of the antiderivative at (b) and (a.)

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So the second part of the fundamental theorem says that if we take a function (F,) first differentiate it, and then integrate the result, we arrive back at the original function, but in the form (Fleft( b right) – Fleft( a right).)

Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes.

The Area under a Curve and between Two Curves

The area under the graph of the function (fleft( x right)) between the vertical lines (x = a,) (x = b) (Figure (2)) is given by the formula

[S = intlimits_a^b {fleft( x right)dx} = {Fleft( b right) – Fleft( a right).}]

Let (Fleft( x right)) and (Gleft( x right)) be antiderivatives of functions (fleft( x right)) and (gleft( x right),) respectively.

If (fleft( x right) ge gleft( x right)) on the closed interval (left[ {a,b} right],) then the area between the curves (y = fleft( x right),) (y = gleft( x right)) and the lines (x = a,) (x = b) (Figure (3)) is given by

[
{S = intlimits_a^b {left[ {fleft( x right) – gleft( x right)} right]dx} }
= {Fleft( b right) – Gleft( b right) }-{ Fleft( a right) + Gleft( a right).}
]

The Method of Substitution for Definite Integrals

The definite integral (intlimits_a^b {fleft( x right)dx} ) of the variable (x) can be changed into an integral with respect to (t) by making the substitution (x = gleft( t right):)

[{intlimits_a^b {fleft( x right)dx} }={ intlimits_c^d {fleft( {gleft( t right)} right)g’left( t right)dt} .}]

The new limits of integration for the variable (t) are given by the formulas

[{c = {g^{ – 1}}left( a right),;;}kern-0.3pt{d = {g^{ – 1}}left( b right),}]

where ({g^{ – 1}}) is the inverse function to (g,) that is (t = {g^{ – 1}}left( x right).)

Integration by Parts for Definite Integrals

In this case the formula for integration by parts looks as follows:

[{intlimits_a^b {udv} }={ left. {uv} right|_a^b – intlimits_a^b {vdu} ,}]

where (left. {uv} right|_a^b) means the difference between the product of functions (uv) at (x = b) and (x = a.)

Solved Problems

Click or tap a problem to see the solution.

Example 1

Calculate the derivative of the function (gleft( x right) = intlimits_1^x {sqrt {{t^3} + 4t} dt} ) at (x = 2.)

Example 2

Calculate the derivative of the function (gleft( x right) = intlimits_{ – large{frac{pi }{2}}normalsize}^x {sqrt {{{sin }^2}t + 2} dt} ) at (x = large{large{frac{pi }{6}}normalsize}.)

Example 3

Find the derivative of the function (gleft( x right) = intlimits_3^{{x^2}} {large{frac{{dt}}{t}}normalsize}.)

Example 4

Find the derivative of the function (gleft( x

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