The Definite Integral

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Lesson Priority: VIP Knowledge

Calculus $\longrightarrow$
Understanding Integration $\longrightarrow$
 
Objectives
  • Learn what a definite integral represents conceptually and visually
  • Split or combine integrals of a function using given information and integration limits
  • Know what happens when you switch the limits of integration and why
  • Relate functions and their Antiderivatives using the a Fundamental Theorem of Calculus
  • Use the FTC to evaluate definite integrals
Lesson Description

This lesson more precisely defines the operation of finding the area under a curve between two $x$ values. We'll look at properties of definite integrals, and relate definite integrals to antiderivatives using a very important theorem.

 
Practice Problems

Practice problems and worksheet coming soon!

 

Definite Integrals and Area

You Should Know
This lesson goes hand-in-hand with the lesson on the Fundamental Theorem of Calculus ». In fact, some teachers will combine these two completely, so your best bet for quiz and test prep is to learn and practice these two lessons' worksheets together.
Similar to how the derivative of a function defines its ever-changing slope, the integral of a function defines the area between the function and the $x$ axis.In order to find bounded area, however, we need to know where to start and stop.We will refer to the vertical boundaries of the area as limits of integration. When we measure specific area of a function, we will put the left and right limits below and above the integral symbol.
Definite Integral NotationLet $f(x)$ be a differentiable function. The area bound between the function, the $x$ axis, and the vertical lines $x=a$ and $x=b$ is notated by$$\int_{a}^{b} f(x) \, dx$$for $a < b$.
Recall also that area above the $x$-axis is considered positive area, while area below the $x$-axis is considered negative.As we'll see, sometimes we will combine areas together, and negative area can "net out" against positive area to reduce the answer, not add to it.
You Should Know
It may seem odd that we have two kinds of integration - indefinite which we recently discussed as antiderivatives with a $+C$ in the answer, and definite integration which we are defining as a specified area. Both types have their use, and you'll always know which is needed because definite integrals will have limits with the integral symbol, while indefinite ones will not.

Technical Definition

Like most things in life, we won't often care about the technical definition. However, you will probably see it in a classroom lecture setting, and if you're taking an AP exam, you're expected to at least be able to recognize it.A definite integral is a limit of a Riemann sum », which we recently studied. Recall that Riemann sums are finite approximations of area based on the rectangle method of approximating the area between a curve and the $x$-axis. The idea was that the more rectangles you use, the better the approximation (although more work).If $n$ represents the number of rectangles we'll use for the approximation, then a Riemann sum looks like$$\sum_{k=1}^{n} f(x^{*}_{k}) \, \Delta x_{k}$$Where $f(x)$ is the function, $x^{*}_{k}$ is the $x$ value of the $k$-th rectangle start point, and $\Delta x_{k}$ is the fixed width of the $k$ rectangles.If we allow an infinite number of rectangles, then we are seeking the limit as $n \to \infty$:$$ \lim_{n \to \infty} \sum_{k=1}^{n} f(x^{*}_{k}) \, \Delta x_{k}$$And, loosely speaking, if the first value of $x^{*}_{k}$ is $a$, and the last value is $b$, then we are calculating the area between $x=a$ and $x=b$ that is trapped between the function and the $x$ axis, leaving us with a result we should be familiar with.$$\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x^{*}_{k}) \, \Delta x_{k}$$
You Should Know
You may be asked to reproduce this limit definition of a definite integral on an AP exam, though more than likely, you would be asked to interpret it in a multiple choice setting.
 
Example 1Which of the following is equal to $\displaystyle \int_{2}^{6} x^3 \, dx$ ?(A)    $\displaystyle \lim_{n \to \infty} \sum_{i=1}^{n} \left( \frac{6i}{n} \right)^3 \cdot \frac{4}{n}$(B)    $\displaystyle \lim_{n \to \infty} \sum_{i=1}^{n} \left( 4 + \frac{2i}{n} \right)^3 \cdot \frac{2}{n}$(C)    $\displaystyle \lim_{n \to \infty} \sum_{i=1}^{n} \left( 2 + \frac{4i}{n} \right)^3 \cdot \frac{4}{n}$(D)    $\displaystyle \lim_{n \to \infty} \sum_{i=1}^{n} \left( 4 + \frac{4i}{n} \right)^3 \cdot \frac{2}{n}$$\blacktriangleright$ For starters, we know the right-most multiplier should represent the rectangle width $\Delta x_{k}$, which is the length of the integration interval ($4$) divided by the number of approximating rectangles ($n$). Therefore, we know right away that the correct answer is either (A) or (C).Recall also that the expression for $x^{*}_{k}$ is $a + i \cdot \Delta x_{k}$, and since we just agreed that $\Delta x_{k}$ is equal to $4/n$, and $a$ (which is the lower integration limit) should be $2$, the correct answer has to be (C).
 

Geometry Questions

Because definite integrals represent the area between the function and the $x$-axis, certain integrals are solvable using only geometry knowledge. If the function graph uses straight lines, we end up with rectangles and triangles and therefore can calculate the areas using geometry formulas.
 
Example 2Using geometry, calculate $\displaystyle \int_{-3}^{1} x \, dx$.$\blacktriangleright$ By now you're probably familiar with this function - a straight line passing through the origin. We can sketch the area described by this integral and calculate the exact amount of area with the formula for the area of a triangle.The area between $x=-3$ and $x=0$ is negative area, because it is below the $x$ axis. The area from $x=0$ to $x=1$ is positive area, because it is above the $x$ axis. For this reason, and because it's easier, we should calculate these two triangle-shaped areas separately.The formula for the area of a triangle is one-half base times height. Using the graph we can see that the lower triangle has an area of$$\frac{1}{2} \cdot 3 \cdot 3 = \frac{9}{2}$$while the upper triangle has an area of$$\frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$$Again, the lower triangle represents negative area, while the upper one represents positive area, so the net total area of this integral is$$- \, \frac{9}{2} + \frac{1}{2} = -4 \;\; \blacksquare$$
 
Let's try another one.
 
Example 3Using geometry, calculate $\displaystyle \int_{-3}^{2} |2x + 1| \, dx$.$\blacktriangleright$ This function has a v-shaped graph, and the area bound between the function and the $x$-axis between $x=-3$ and $x=2$ consists of two triangles, each of which we can quickly find the area of.The first triangle is $\displaystyle 2 \, \frac{1}{2}$ units wide and $5$ units tall.$$\mathrm{Area} = \frac{1}{2} \cdot \frac{5}{2} \cdot 5 $$$$=\frac{25}{4}$$The second triangle is also $\displaystyle 2 \, \frac{1}{2}$ units wide, and $5$ units tall, so its area will also be $25/4$.Both areas are above the $x$ axis, so they are each positive quantities. Overall, that means the total area is $25/4 + 25/4$ or $25/2$ square units. Hence,$$\int_{-3}^{2} |2x + 1| \, dx = \frac{25}{2} \; \blacksquare$$
 

Splitting and Combining Limits

As we begin to work with the definite integral as a concept, we need to learn a few characteristics that are specific to the idea of integration limits, starting with splitting and combining limits.
You Should Know
These two concepts are simple but powerful. Most teachers use these ideas heavily in the tests on this unit.
Take a look at the following picture of an area represented by the integral of $f(x)$ from $a$ to $c$:If there was some in-between value $b$ that divided this area into two pieces, we can represent each piece of area as its own integral.Also, it makes sense that the sum of each piece would be equal to the entire original area. This picture representation of combining integrals is one way to informally prove the following theorem.
Combining IntegralsLet $f(x)$ be a continuous function over an interval, and let $a$, $b$, and $c$ be in that continuous interval. It follows that$$\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx$$
Similarly, we can state this idea backward to prove that, if it is convenient to do so, we can split an integral into pieces.
Splitting IntegralsLet $f(x)$ be a continuous function over an interval, and let $a$, $b$, and $c$ be in that continuous interval. It follows that$$\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx$$
 
Example 4Which of the following three statements is guaranteed to be true?(A)    $\displaystyle \int_{1}^{3} f(x) \, dx = \int_{1}^{2} f(x) \, dx + \int_{1}^{2} \, dx$(B)    $\displaystyle \int_{0}^{5} f(x) \, dx = \int_{0}^{6} f(x) \, dx - \int_{5}^{6} f(x) \, dx$(C)    $\displaystyle \int_{-2}^{2} f(x) \, dx = 2 \int_{0}^{2} f{x} \, dx$
Show solution
$\blacktriangleright$ (B) is correct. Let's see why.The integral from $0$ to $6$ can be split up into two integrals: one from $0$ to $5$ and another from $5$ to $6$.Therefore we have$$\int_{0}^{6} f(x) \, dx - \int_{5}^{6} f(x) \, dx$$$$=\Bigg[ \int_{0}^{5} f(x) \, dx + \int_{5}^{6} f(x) \, dx \Bigg] - \int_{5}^{6} f(x) \, dx$$$$= \int_{0}^{5} f(x) \, dx \;\; \blacksquare$$
 
To be clear, the last example also proved that we can subtract away part of an integral to obtain a shorter one.
Subtracting IntegralsLet $a$, $b$, and $c$ be $x$ values. For a continuous function $f(x)$, we have$$\int_{a}^{b} f(x) \, dx - \int_{a}^{c} f(x) \, dx = \int_{c}^{b} f(x) \, dx$$
In a way, this is a straight-forward adaptation of the combination principle we just defined. Also, this works for any $a$, $b$, or $c$, and in fact it does not matter which number is larger, due to what we're about to see about switching the order of integration limits.

Switching Integration Limits

Another integration limits rule we need to understand is the relationship between two integrals with the same limits but in reverse order. This rule is infrequently useful and not very intuitive, but when it surfaces we do need to know it.
Switching Integration LimitsThe integral from $a$ to $b$ of a function is equal to the negative of the integral from $b$ to $a$ of the same function.$$\int_{a}^{b} f(x) \, dx = - \int_{b}^{a} f(x) \, dx$$
This isn't intuitive and many students do not like it, because that means one of the integrals is "backward" since either $a$ or $b$ must be the bigger of the two numbers. The geometric interpretation of this phenomenon is that measuring the same area from left to right is the negative equivalent result of measuring from right to left:Again, we don't often have to deal with this phenomenon, and when we do, it's almost always algebraic.
 
Example 5Which of the following three statements is guaranteed to be true?(A)    $\displaystyle \int_{1}^{3} f(x) \, dx = \int_{-1}^{-3} f(x) \, dx$(B)    $\displaystyle \int_{-3}^{5} f(x) \, dx = -\int_{-5}^{3} f(x) \, dx$(C)    $\displaystyle \int_{2}^{4} f(x) \, dx = -\int_{4}^{2} f(x) \, dx$
Show solution
$\blacktriangleright$ Answer choice (C) is the correct expression, based on the limit-switching rule. Neither (A) nor (B) follows the limit switch rule, nor any other integration limits rule. $\blacksquare$
 
Pro Tip
The fact that switching integration limits is equivalent to negating the integral is not an intuitive property, nor is it often useful. It's one of those oddly specific properties that we need to know for quizzes, and be able to understand when it shows up infrequently in the future.

Linear Operator Properties

We have discussed linear operator properties in the course already, first for derivative operations » and again recently for introductory integration » concepts.Integrals of SumsFirst is the idea that we can split an integral of a sum into the sum of integrals. This was discussed for general antiderivatives recently but the same holds true for definite integrals, as long as the limits stay the same.That is,$$\int_{1}^{5} x^2 + 2x \, dx = \int_{1}^{5} x^2 \, dx + \int_{1}^{5} 2x \, dx$$but it wouldn't stay true if the limits changed:$$\int_{1}^{5} x^2 + 2x \, dx \ne \int_{4}^{8} x^2 \, dx + \int_{4}^{8} 2x \, dx$$Constants and IntegralsConstant coefficients can "come outside" the integral. For example,$$\int_{0}^{4} 2x^3 \, dx = 2 \int_{0}^{4} x^3 \, dx$$In the context of area, this makes a lot of sense. If a particular definite integral equals some area, then the integral of two times the same function should give you two times as much area, since $2f(x)$ is twice as tall as $f(x)$.Also, the integral of a constant on its own is equal to the constant multiplied by $x$. The upcoming lesson on polynomial antiderivatives » will reinforce why this is the case.This is another intuitive idea when applied to area with definite integrals, because if $f(x)$ is a constant, then its graph is a horizontal line. This means that the area between the function and the $x$-axis is a rectangle, and we could justify this result with geometry.
 
Example 6Calculate $\displaystyle \int_{2}^{5} 6 \, dx$ by using geometry.$\blacktriangleright$ The function $f(x) = 6$ is a horizontal line six units above the $x$-axis. Integrating this function from $2$ to $5$ means we should start at $x=2$ and stop at $x=5$.This enclosed area is a rectangle with height $6$ and width $3$. Therefore the area is $18$. You may not need a picture to see why but I always encourage a quick sketch if it is helpful to you.
 

Unknown Function Questions

If you are given values of the integrals of unknown functions, you can calculate various combinations of related integrals with different limits.
 
Example 7Given the following information:$$\int_{0}^{5} f(x) \, dx = 3$$$$\int_{-1}^{3} f(x) \, dx = -2$$$$\int_{-5}^{3} f(x) \, dx = 4$$$$\int_{3}^{5} f(x) \, dx = 6$$$$\int_{4}^{5} f(x) \, dx = -1$$Find the values of the following quantities by using the properties of combining and splitting integrals:(a) $\displaystyle \int_{-5}^{-1} f(x) \, dx$
Show solution
$$\blacktriangleright \; \int_{-5}^{3} f(x) \, dx = \int_{-5}^{-1} f(x) \, dx + \int_{-1}^{3} f(x) \, dx$$$$4 = \int_{-5}^{-1} f(x) \, dx + (-2)$$$$\int_{-5}^{-1} f(x) \, dx = 4 + 2 = 6 \;\; \blacksquare$$
(b) $\displaystyle \int_{0}^{4} f(x) \, dx$
Show solution
$$\blacktriangleright \; \int_{0}^{5} f(x) \, dx = \int_{0}^{4} f(x) \, dx + \int_{4}^{5} f(x) \, dx$$$$3 = \int_{0}^{4} f(x) \, dx + (-1)$$$$\int_{0}^{4} f(x) \, dx = 3 + 1 = 4 \;\; \blacksquare$$
(c) $\displaystyle \int_{3}^{4} f(x) \, dx$
Show solution
$$\blacktriangleright \; \int_{3}^{5} f(x) \, dx = \int_{3}^{4} f(x) \, dx + \int_{4}^{5} f(x) \, dx$$$$6 = \int_{3}^{4} f(x) \, dx + (-1)$$$$\int_{3}^{4} f(x) \, dx = 6 + 1 = 7 \;\; \blacksquare$$
(d) $\displaystyle \int_{-5}^{3} f(x) \, dx$
Show solution
$$\blacktriangleright \; \int_{-5}^{5} f(x) \, dx = \int_{-5}^{3} f(x) \, dx + \int_{3}^{5} f(x) \, dx$$$$\int_{-5}^{5} f(x) \, dx = 4 + 6 = 10 \;\; \blacksquare$$
 
You Should Know
This is a very popular quiz question because there are no graphs or function definitions, which can confuse students who didn't study!

Mr. Math Makes It Mean

Odd FunctionsOn occasion, the definite integral concept is the only by-hand solution to integrals that would otherwise require a calculator. Specifically, any odd function will have a net area of zero when integrated from $-k$ to $k$, because of the symmetry properties of odd functions.Recall that a function is odd if$$f(-x) = -f(x)$$A function like $\displaystyle f(x) = \sin\left( x^3 \right)$ is odd because$$f(-x) = \sin\left( (-x)^3 \right) = \sin\left( -x^3 \right)$$and since $\sin$ is an odd function,$$\sin\left( -x^3 \right) = -\sin \left (x^3 \right) = -f(x)$$.Therefore, when we're asked to compute something like$$\int_{-2}^{2} \sin \left( x^3 \right) \, dx$$without a calculator, we'll know the answer is zero.Zero Width IntegralsIf you ever come across an integral with limits that are the same number, the answer is automatically zero. This is the same as trying to integrate an area with zero width. Teachers sometimes put these trick integral limits in front a function that actually can't be integrated, like$$\int_{1}^{1} \sin \left( e^{x^2} \right) \, dx$$The Fundamental Theorem of Calculus » also reinforces why any integral with identical start and stop limits should equal zero. This is a ridiculous trick in my humble opinion but the good news is that if your teacher tries to fool you with this idea, it's not complicated to understand and answer.
 

Put It To The Test

Example 8Which of the following represents $\displaystyle \int_{-4}^{4} x^2 \, dx$$$\mathrm{(a)} \;\;\;\; \lim_{n \to \infty} \sum_{i=1}^{n} \left( -4 + \frac{4i}{n} \right)^2 \cdot \frac{8}{n}$$$$\mathrm{(b)} \;\;\;\; \lim_{n \to \infty} \sum_{i=1}^{n} \left( -4 + \frac{8i}{n} \right)^2 \cdot \frac{8}{n}$$$$\mathrm{(c)} \;\;\;\; \lim_{n \to \infty} \sum_{i=1}^{n} \left( -4 + \frac{4i}{n} \right)^2 \cdot \frac{4}{n}$$$$\mathrm{(d)} \;\;\;\; \lim_{n \to \infty} \sum_{i=1}^{n} \left( \frac{64i^2}{n^2} \right)^2 \cdot \frac{8}{n}$$
Show solution
$\blacktriangleright$ Recall that the Riemann sum uses $\Delta x_{k}$ as the multiplier on the end:$$\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f \left( x^{*}_{k} \right) \cdot \Delta x_{k}$$where $\displaystyle \Delta x_{k} = \frac{(b-a)}{n}$ and $\displaystyle x^{*}_{k} = a + i \cdot \Delta x_{k}$.Therefore, we're looking for $8/n$ at the end of the expression. Also, we have$$x^{*}_{k} = -4 + \frac{8i}{n}$$and so$$f \left( x^{*}_{k} \right) = \left( -4 + \frac{8i}{n} \right)^2$$leading us to answer choice (b). $\blacksquare$
 
Example 9Sketch the region described by the following definite integral, and then evaluate the integral using geometry knowledge.$$\int_{0}^{6} |x-4| \, dx$$
Show solution
$\blacktriangleright$ Your sketch doesn't need to be absolutely perfect but good enough for you to figure out the shapes and measures that you'll need to apply the basic geometry knowledge - in this case, the formula for the area of a triangle.Both triangles are above the $x$ axis, so the net area of each will be positive.The left-most triangle has a base length of $4$ and a height of $4$, giving an area of $8$. The right-most triangle has a base of $2$ and a height of $2$, giving an area of $2$.Therefore, the result of this definite integral is $8$ + $2$, or $10$. $\blacksquare$
 
Example 10Write the following integral as the sum of two integrals with different limits.$$\int_{-3}^{6} g(x) \, dx$$
Show solution
$\blacktriangleright$ There are several ways to do this. Most students will split this into the sum of two integrals by picking a random number in between $-3$ and $6$, as this is a natural and relatively simple solution. e.g.$$\int_{-3}^{6} g(x) \, dx = \int_{-3}^{1} g(x) \, dx + \int_{1}^{6} g(x) \, dx$$It's also possible to write this as the difference of two integrals. Most students would say it seems more complicated to do this, but it's just as fine in terms of a correct solution. e.g.$$\int_{-3}^{6} g(x) \, dx = \int_{-3}^{10} g(x) \, dx - \int_{6}^{10} g(x) \, dx \;\; \blacksquare$$
 
Example 11Simplify.$$\int_{-3}^{6} f(x) - g(x) \, dx$$$$+\int_{-5}^{-3} f(x) \, dx$$$$+\int_{0}^{6} g(x) \, dx$$
Show solution
$\blacktriangleright$ You can hash this out using the formula technically but it's much easier to think conceptually, focusing on each function.For function $f(x)$, we are integrating from $-3$ to $6$, and then adding the integral from $-5$ to $-3$. All in all, this means we are integrating $f$ from $-5$ to $6$.$$\int_{-3}^{6} f(x) \, dx + \int_{-5}^{-3} f(x) \, dx = \int_{-5}^{6} f(x) \, dx$$For function $g(x)$, we are integrating from $0$ to $6$ and subtracting the integral from $-3$ to $6$. This is a little more confusing, but you can split the integrals to help you see the answer:$$\int_{0}^{6} g(x) \, dx - \int_{-3}^{6} g(x) \, dx$$$$= \int_{0}^{6} g(x) \, dx - \left[ \int_{-3}^{0} g(x) \, dx + \int_{0}^{6} g(x) \, dx \right]$$$$= -\int_{-3}^{0} g(x) \, dx$$Putting both function results together, the answer is$$\int_{-5}^{6} f(x) \, dx - \int_{-3}^{0} g(x) \, dx \;\; \blacksquare$$
 
Example 12Which of the following four statements are always true?$$\begin{align} \mathrm{I.} & \;\;\;\; \int_{-2}^{0} f(x) \, dx = -\int_{0}^{-2} f(x) \, dx \\ \mathrm{II.} & \;\;\;\; \int_{-5}^{5} f(x) \, dx = \int_{-5}^{-1} f(x) \, dx + \int_{-1}^{5} f(x) \, dx \\ \mathrm{III.} & \;\;\;\; \int_{0}^{3} f(x) \, dx > \int_{0}^{1} f(x) \, dx \\ \mathrm{IV.} & \;\;\;\; -\int_{a}^{b} f(x) \, dx = \int_{a}^{b} -f(x) \, dx \end{align}$$
Show solution
$\blacktriangleright$ I, II, and IV are correct. Statement I correctly states the rule about what happens when you switch the limits of integration, statement II is a correct application of splitting an integral into two pieces, and statement IV is correct because it shows a constant (in this case $-1$) being factored out of the integral, which is allowed by the linear operator properties of integration.Statement III is incorrect because it is a statement that can be true or false depending on what $f(x)$ is. $\blacksquare$
 
Example 13Use the properties of linear operators to write the following expression in terms of separate integrals of only $f(x)$, $g(x)$, and $h(x)$.$$\int_{0}^{10} 2f(x) - 3g(x) + 3h(x) \, dx$$
Show solution
$\blacktriangleright$ Linear operator properties tell us that we can (1) split up the integrals into three integrals, and (2) factor coefficients out of the integral, keeping them in front. Ultimately you should end up writing:$$2 \int_{0}^{10} f(x) \, dx - 3 \int_{0}^{10} g(x) \, dx + 3 \int_{0}^{10} h(x) \, dx \;\; \blacksquare$$
 
Example 14Given:$\displaystyle \int_{0}^{4} f(x) \, dx = 5$$\displaystyle \int_{0}^{1} f(x) \, dx = 2$$\displaystyle \int_{1}^{5} f(x) \, dx = 4$$\displaystyle \int_{0}^{4} g(x) \, dx = -4$$\displaystyle \int_{1}^{4} g(x) \, dx = -3$Find the values of each of the following:(a) $\displaystyle \int_{1}^{4} f(x) - g(x) \, dx$(b) $\displaystyle \int_{0}^{1} f(x) + g(x) \, dx$(c) $\displaystyle \int_{4}^{5} f(x) \, dx$
Show solution
$\blacktriangleright$ For part (a), we already have what we need for $g(x)$. Obtain the needed $f$ integral by using the first two pieces of information:$$\int_{1}^{4} f(x) \, dx = \int_{0}^{4} f(x) \, dx - \int_{0}^{1} f(x) \, dx$$$$ = 5 - 2 = 3$$Altogether, $\displaystyle \int_{1}^{4} f(x) - g(x) \, dx = (3) - (-3) = 6$For part (b), we have what we need for $f(x)$ but need to calculate the $g$ integral. The last two pieces of information can help:$$\int_{0}^{1} g(x) \, dx = \int_{0}^{4} g(x) \, dx - \int_{1}^{4} g(x) \, dx$$$$ = -4 - (-3) = -1$$Overall, we have$$\int_{0}^{1} f(x) + g(x) \, dx = 2 + (-1) = 1$$For part (c), the easiest way to calculate this integral is$$\int_{4}^{5} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{5} f(x) \, dx - \int_{0}^{4} f(x) \, dx$$(hint: this is a good one to try to explain to yourself why in words)This is equal to $2 + 4 - 5$, or $1$. $\blacksquare$
 
Lesson Takeaways
  • Understand definite integrals as a concept of measuring area between a function and the $x$ axis
  • Be able to recognize the Riemann sum form of a given definite integral
  • Know how to find definite integrals of unknown functions using their graphs and geometry
  • Be familiar with linear operator concepts, as well as rules for splitting and combining integrals

Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

Key Lesson Sections

Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.

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