If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Class 12 math (India)

Course: Class 12 math (India)>Unit 10

Lesson 4: Definite integral as area

Worked example: Definite integral by thinking about the function's graph

If you already know the area under a curve, you can use it to compute an integral.

Want to join the conversation?

• When we solve definite integral, do we always have to graph the function?
Also, why does definite integral look like the Anti-derivative symbol?
(8 votes)
• No, we often solve definite integrals without graphing them, although graphing is often helpful in understanding the problem and checking to see if our result is reasonable. As for the symbols, it may seem odd that we use the same one for two entirely different concepts, one of which yields a function while the other yields a constant -- but the relationship between these two concepts is so important that its proof is known as the Fundamental Theorem of Calculus. It is our ability to use antiderivatives to solve definite integrals that gives calculus much of its power.
(37 votes)
• where are the intro videos of definite integral he started this from simply evaluating it
(11 votes)
• how does he know that it just includes the top part of the circle?
(6 votes)
• The equation of a circle centred at the origin with radius `3` is `x² + y² = 9`. Hence `y² = 9 - x²`. But then we have `y = √(9 - x²)`, or we have `y = -√(9 - x²)`. Since he is dealing with `√(9 - x²)` in the video, we know that he is dealing with points on a circle with `y ≥ 0`. Hence only the "top" of the circle.
(15 votes)
• Why is the radius the square root of x^2 + y^2?
(2 votes)
• The Pythagorean theorem states that `x² + y² = r²`. Now take the square root of this.

On a deeper level, that is the way we compute distance in `R²`, using the Euclidean norm, or the Euclidean metric, but you do not have to worry about such things as a normed linear space or a metric space at the moment. Go with the short answer at the top...
(13 votes)
• so integral means finding the area below the graph ?
(2 votes)
• That is one use for integrals. There are other uses.
(6 votes)
• Hey I just wonder why we use the integral notation in indefinite integral rather than directly use f(x)dx ? It makes sense when we use the "s"notation from a to b in definite integral which I regard it's meaning is same as the "sum" of f(x)dx when there is a infinite point from a to b
(3 votes)
• This is a keen observation. We have the elongated "s" without the upper and lower bounds ( b and a respectively) for indefinite integrals because they are actually related to definite integrals. The indefinite integral is defined as the area under a function f(u) from u=a ( a being a constant) to u=x (x being a variable). The definite integral is a function of the upper and lower bounds, you can interpret the indefinite integral as being a function of just the upper bound, ( we are holding the lower bound a as constant and making the upper bound b vary which is denoted by changing b to x)

We omit the upper bound u=x and the lower bound u=a to keep things simple.
(3 votes)
• If we HAD been taking the definite integral of +/- sqrt(9-x^2), and the graph had just been a circle, would it have summed to zero?
(2 votes)
• You could do this as two separate functions, taking the definite integral of +sqrt(9-x^2) and -sqrt(9-x^2), then adding the two results at the end. This would indeed equal zero, but only if the circle is centered around the origin.
(3 votes)
• I don't know to graph functions.I have to begin from the scratch.Where should I start?Please provide me links if possible!
(1 vote)
• (3 votes)
• Why didnt it end up as a line of `3-x`, since `√9 = 3` and `√x^2 =x` ?
(1 vote)
• Sal wants us to evaluate the definite integral of (9-x^2)^1/2 from -3 to 3. Instead he simply manipulates the function y= (9-x^2)^1/2 dx, takes the square of both sides and simply forgets the dx. I thought to calculate the definite integral of a function we first had to determine the function derived from and then use our interval values while dropping the constant C as this would not be an indefinite integral.
(2 votes)

Video transcript

- [Voiceover] What I want to do in this video is see if we can evaluate the definite integral from negative three to three of the square root of nine minus x squared dx. I encourage you to pause this video and try it on your own. I'll give you a hint, you can do this purely by looking at the graph of this function. All right, I'm assuming you've had a go at it. So let's just think about, I just told you that you could do this by using the graph of the function. Let's graph this function. Let's get a y-axis here. This is my y-axis. This is my x-axis. You might be saying, "Oh well, what is the graph "of this thing?", it might not jump out at you. It's been a little while (mumble) of a hint since you've done conic sections, maybe in your Algebra class. Let's just remind ourselves. If this function, if we said y is equal to some function of x, which we see is the square root of nine minus x squared. Then we could say, "Well, that means that y squared must "must be equal to this thing squared." Which is nine minus x squared. Then we could say, "Y squared plus x squared is equal "to nine." and you might recognize this as a circle centered at origin with radius equal to three, the square root of nine. So radius is equal to three, centered at the origin. Now, they graph of this is not going to be a circle. This is a function. It would be a circle and it would not be a function anymore if you said the positive and negative square roots of here. When we took the squares of both sides we got the bottom back I guess you could say. But up here, we're only talking about the principal root. When you're talking about the principal root you're really talking about the top. This is the top of a circle centered at the origin with radius three. So this is top of circle cause it's the positive square root, so let's draw that. It's gonna have a radius of three, centered at the origin. This is gonna be negative three, this is going to be three. This is going to be three right over here. So this function is going to look like this. It's actually only defined between negative three and three. The absolute value of x is greater than three, then you're going to get a negative value in here. Then you can't take the principal root, as if we're defining it over a positive or non-negative values I should say. So this is the graph. What is the definite integral from negative three to three. Well it's just the area under the curve and above the x-axis, it's the stuff that I am shading in in green. Well, what's that? We don't need Calculus to figure that out. You can do this with just traditional geometry. The area of the entire circle, if there were an entire circle, would just be pi r squared. So it would be pi times three squared, which is equal to nine pi. Now this is only half of the entire circle. So we're gonna divide that by two. The area is nine pi over two. So this thing is nine pi over two.