Concept of an integral, areas with definite integrals, and basic anti-derivative solving skills
Integrals are a powerful tool that allow us to calculate the area under a curve.
The essence of integration is to draw rectangles under the curve, each with a height matching the function at each point. Adding up the area of these rectangles gives an approximation for the area under the curve. Increasing the number of rectangles makes the approximation more precise.
The real "trick" of integration is imagining what happens when the number of rectangles approaches infinity. In this case, the width each each rectangle approaches zero, and their combined area approaches that of the true region under the curve.
This infinite sum of rectangles is what we call a definite integral, and uses the notation
The word definite here refers to the boundsĀ āaāĀ andĀ āb.Ā Later in this lesson we'll explain what an integral without bounds might mean.
The area of an infinite number of rectangles might seem a strange concept. But in reality, by following careful mathematical steps we can actually find the exact area under most curves. This process is called Riemann Integration, and the IB does not expect you to know how to perform it exactly. Instead, we will learn a series of rules for calculating
The area between a curve āf(x)>0ā and theĀ āxā-axis is given by
The trapezoid rule is a technique for more accurately approximating the area beneath a curve fromĀ āx=aāĀ toĀ āx=b.Ā Instead of summing rectangles, it works by summing the area ofĀ ānāĀ trapezoids of equivalent width.
Each trapezoid's area is the common widthĀ ā(nbāaā)āĀ times the average of the function's value on the left and right side of the trapezoid:
The formula for the area using the trapezoid rule approximation is
whereĀ āh=nbāaā.
Recognize thatĀ āb=3,Ā Ā āa=0āĀ andĀ ān=6,Ā so the width of each trapezoid is
Using the formula, the area is
EachĀ āykā=e0.5Ćk,Ā but thankfully we can use our calculator here:
InĀ āL1āāĀ enterĀ ā0,1ā¦6.
At the top ofĀ āL2ā,Ā setĀ āL2ā=e0.5L1āāĀ and hit enter. The values fill in:
NowĀ ā(y0ā+y6ā+2(y1ā+āÆ+y5ā))āĀ is
So the area is
Integration, or anti-differentiation, is essentially the opposite of differentiation. We use the integral symbolāā«āĀ and write:
By convention we denote this functionĀ āF:
We can also write
Notice theĀ ādxāĀ under the integral. This tells us which variable we are integrating with respect to - in this case we are reversingĀ ādxdā.
Since the derivative of a constant is always zero, then if ifĀ āFā²(x)=f(x),Ā thenĀ ā(F(x)+C)ā²=f(x).
This means that when we integrate, we can add any constant to our result, since differentiating makes this constant irrelevant:
In the same way that constant multiples can pass through the derivative, they can pass through the integral:
And in the same way that the derivative of a sum is the sum of the derivatives:
If we know the value ofĀ āyāĀ orĀ āf(x)āĀ for a givenĀ āx,Ā we can determineĀ āCāĀ by plugging inĀ āxāĀ andĀ āy.
A definite integral is evaluated between a lower and upper bound.
We can solve a definite integral with
whereĀ āF(x)=ā«f(x)dx.
Graphing calculators can be used to evaluate definite integrals.
For example, on a TI-84, math > 9:fnInt(, which prompts you withĀ āā«ā”ā”ā(ā”)dā”.Ā Make sure the variable of your function matches the variable that you take the integral with respect to.
Integrals of the same function with adjacent domains can be merged:
Similarly, the domain of an integral can be split:
for anyĀ āa<m<b.
In general, the area enclosed between a curve and theĀ āxā-axis is given by
since any region below theĀ āxā-axis hasĀ āf(x)<0,Ā but area must always be positive.
This can be done with technology, or by splitting the integral into parts - whereĀ āfāĀ is positive and whereĀ āfāĀ is negative:
The area enclosed between two curves is given by
This can be done with technology, or by splitting the integral into multiple regions, each having eitherĀ āf(x)>g(x)āĀ orĀ āg(x)>f(x).
Nice work completing Definite Integrals and Basic Anti-Derivatives, here's a quick recap of what we covered:
Exercises checked off
Concept of an integral, areas with definite integrals, and basic anti-derivative solving skills
Integrals are a powerful tool that allow us to calculate the area under a curve.
The essence of integration is to draw rectangles under the curve, each with a height matching the function at each point. Adding up the area of these rectangles gives an approximation for the area under the curve. Increasing the number of rectangles makes the approximation more precise.
The real "trick" of integration is imagining what happens when the number of rectangles approaches infinity. In this case, the width each each rectangle approaches zero, and their combined area approaches that of the true region under the curve.
This infinite sum of rectangles is what we call a definite integral, and uses the notation
The word definite here refers to the boundsĀ āaāĀ andĀ āb.Ā Later in this lesson we'll explain what an integral without bounds might mean.
The area of an infinite number of rectangles might seem a strange concept. But in reality, by following careful mathematical steps we can actually find the exact area under most curves. This process is called Riemann Integration, and the IB does not expect you to know how to perform it exactly. Instead, we will learn a series of rules for calculating
The area between a curve āf(x)>0ā and theĀ āxā-axis is given by
The trapezoid rule is a technique for more accurately approximating the area beneath a curve fromĀ āx=aāĀ toĀ āx=b.Ā Instead of summing rectangles, it works by summing the area ofĀ ānāĀ trapezoids of equivalent width.
Each trapezoid's area is the common widthĀ ā(nbāaā)āĀ times the average of the function's value on the left and right side of the trapezoid:
The formula for the area using the trapezoid rule approximation is
whereĀ āh=nbāaā.
Recognize thatĀ āb=3,Ā Ā āa=0āĀ andĀ ān=6,Ā so the width of each trapezoid is
Using the formula, the area is
EachĀ āykā=e0.5Ćk,Ā but thankfully we can use our calculator here:
InĀ āL1āāĀ enterĀ ā0,1ā¦6.
At the top ofĀ āL2ā,Ā setĀ āL2ā=e0.5L1āāĀ and hit enter. The values fill in:
NowĀ ā(y0ā+y6ā+2(y1ā+āÆ+y5ā))āĀ is
So the area is
Integration, or anti-differentiation, is essentially the opposite of differentiation. We use the integral symbolāā«āĀ and write:
By convention we denote this functionĀ āF:
We can also write
Notice theĀ ādxāĀ under the integral. This tells us which variable we are integrating with respect to - in this case we are reversingĀ ādxdā.
Since the derivative of a constant is always zero, then if ifĀ āFā²(x)=f(x),Ā thenĀ ā(F(x)+C)ā²=f(x).
This means that when we integrate, we can add any constant to our result, since differentiating makes this constant irrelevant:
In the same way that constant multiples can pass through the derivative, they can pass through the integral:
And in the same way that the derivative of a sum is the sum of the derivatives:
If we know the value ofĀ āyāĀ orĀ āf(x)āĀ for a givenĀ āx,Ā we can determineĀ āCāĀ by plugging inĀ āxāĀ andĀ āy.
A definite integral is evaluated between a lower and upper bound.
We can solve a definite integral with
whereĀ āF(x)=ā«f(x)dx.
Graphing calculators can be used to evaluate definite integrals.
For example, on a TI-84, math > 9:fnInt(, which prompts you withĀ āā«ā”ā”ā(ā”)dā”.Ā Make sure the variable of your function matches the variable that you take the integral with respect to.
Integrals of the same function with adjacent domains can be merged:
Similarly, the domain of an integral can be split:
for anyĀ āa<m<b.
In general, the area enclosed between a curve and theĀ āxā-axis is given by
since any region below theĀ āxā-axis hasĀ āf(x)<0,Ā but area must always be positive.
This can be done with technology, or by splitting the integral into parts - whereĀ āfāĀ is positive and whereĀ āfāĀ is negative:
The area enclosed between two curves is given by
This can be done with technology, or by splitting the integral into multiple regions, each having eitherĀ āf(x)>g(x)āĀ orĀ āg(x)>f(x).
Nice work completing Definite Integrals and Basic Anti-Derivatives, here's a quick recap of what we covered:
Exercises checked off
Beyond the syllabus, it can also help to
Beyond the syllabus, it can also help to