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✍️ Free Response Questions (FRQ)
Calculus Free Response Questions
👑 Unit 1: Limits & Continuity
1.5Determining Limits Using Algebraic Properties of Limits
1.6Determining Limits Using Algebraic Manipulation
1.10Exploring Types of Discontinuities
1.11Defining Continuity at a Point
1.12Confirming Continuity over an Interval
🤓 Unit 2: Differentiation: Definition & Fundamental Properties
2.4Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
🤙🏽 Unit 3: Differentiation: Composite, Implicit & Inverse Functions
3.0Unit 3 Overview: Differentiation: Composite, Implicit, and Inverse Functions
3.1The Chain Rule
3.3Differentiating Inverse Functions
3.4Differentiating Inverse Trigonometric Functions
👀 Unit 4: Contextual Applications of the Differentiation
4.2Straight-Line Motion: Connecting Position, Velocity, and Acceleration
4.4Intro to Related Rates
4.6Approximating Values of a Function Using Local Linearity and Linearization
✨ Unit 5: Analytical Applications of Differentiation
5.0Unit 5 Overview: Analytical Applications of Differentiation
5.2Extreme Value Theorem, Global vs Local Extrema, and Critical Points
5.3Determining Intervals on Which a Function is Increasing or Decreasing
5.4Using the First Derivative Test to Determine Relative (Local) Extrema
5.7Using the Second Derivative Test to Determine Extrema
🔥 Unit 6: Integration and Accumulation of Change
6.11Integrating Using Integration by Parts
💎 Unit 7: Differential Equations
7.0Unit 7 Overview: Differential Equations
7.7Finding Particular Solutions Using Initial Conditions and Separation of Variables
🐶 Unit 8: Applications of Integration
8.1Finding the Average Value of a Function on an Interval
8.2Connecting Position, Velocity, and Acceleration of Functions Using Integrals
8.3Using Accumulation Functions and Definite Integrals in Applied Contexts
8.4Finding the Area Between Curves Expressed as Functions of x
8.5Finding the Area Between Curves Expressed as Functions of y
8.6Finding the Area Between Curves That Intersect at More Than Two Points
8.7Volumes with Cross Sections: Squares and Rectangles
8.8Volumes with Cross Sections: Triangles and Semicircles
8.9Volume with Disc Method: Revolving Around the x- or y-Axis
8.10Volume with Disc Method: Revolving Around Other Axes
8.11Volume with Washer Method: Revolving Around the x- or y-Axis
🦖 Unit 9: Parametric Equations, Polar Coordinates & Vector Valued Functions (BC Only)
9.0Unit 9 Overview: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
9.1Defining and Differentiating Parametric Equations
♾ Unit 10: Infinite Sequences and Series (BC Only)
10.0Unit 10 Overview: Infinite Series and Sequences
10.1Defining Convergent and Divergent Infinite Series
10.6Comparison Tests for Convergence
10.7Alternating Series Test for Convergence
10.1110.11 Finding Taylor Polynomial Approximations of Functions
10.14Finding Taylor or Maclaurin Series for a Function
🧐 Multiple Choice Questions (MCQ)
⏱️ 3 min read
April 26, 2020
Enduring Understanding FUN-6: Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.
Essential Knowledge FUN-6.E.1: Integration by parts is a technique for finding antiderivatives.
Integration by parts is a method of integration that transforms an integral of a product of functions into an integral of the product of one function’s derivative and the others antiderivative. Although this sounds confusing at first, we can build this concept from the product rule: 🥇
Whenever there’s an integral where the integrand (the thing being integrated) is a product of two functions, one of the functions will be f(x) and the other will be g’(x). If other methods like u-substitution don’t work, try integration by parts. 💯
Sometimes, you might see the equation written like this:
This is the same concept in a different form.
One of the functions will be U and the other will be V. However, we still need to know what dV and dU actually mean. If U is a function of x, then U = f(x). Taking the derivative of both sides, we find that dU = f’(x)dx. Same thing for V: If V = g(x), then dV = g’(x)dx. This new way of writing the integration by parts formula is just a sneakier version of the original formula. 🤭
The best way to explain this formula is through an example. Take the following integral:
At first glance, you may try u-substitution, but this method will take you nowhere. Instead, notice that the integrand is a product of two functions. This is an opportunity to use integration by parts. Let’s recall the formula:
The first step should be to figure out what f(x) and g’(x) should be. This step is probably the most challenging of this method, because you have to choose which is which in order to make the problem work.
This is where a handy acronym comes into play: L.I.A.T.E. It stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential.
The function closer to the “E” side should be g’(x) and the function closer to the “L” side should be f(x).
For our example problem, that means:
We know we need g(x) and f’(x) to use the integration by parts formula, so let’s find those now:
Check to make sure that u-substitution and other methods don’t work.
Check to make sure that the integrand is a product of two functions.
Use L.I.A.T.E. to pick which function should be f(x) and which should be g’(x).
Plug into the formula and integrate.
Note: You may have to use integration by parts more than once if your resulting integral is also a product of two functions that can't be solved using u-substitution.
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