Find ∫ 0 6 f ( x ) d x \int_0^6\!f\left(x\right)\,dx ∫ 0 6 f ( x ) d x given the graph of y = f ( x ) y=f\left(x\right) y = f ( x ) .

this problem, we are asked to find the definite integral from zero to six of our function f of x with respect to x given the graph below. Now notice on the graph that we have, we are given this function f of x. It is the curve that we see in red here. And because of this, we're trying to find the area underneath the curve of this function, which is what this definite integral represents. So we're trying to figure out what this area would be underneath this curve, but how do we go about doing that? Well, what I'm going to do is think about some common shapes that I can break this up into in order to calculate this area. Now to do this, there's a couple things that I can do. There's a couple actions I can take here. What I'm first going to do is split this down the middle right about here. Notice this is going to give me a triangle on this side here. Now next, I'm going to split another piece off right here, giving me another triangle as well as a rectangle. These are all shapes that I'm familiar with. So what we'd have is area one plus area two plus area three, and this would give me the total area underneath the curve of this function. Now let's go ahead and start with area one. Area one is gonna be the area of a triangle, which is one half base times height. Now the base of this triangle, well, that's going to be this distance down here, and notice we go from zero all the way over to two. So this base is going to be two. Now the height goes from zero on our y axis all the way up to three. So our height is gonna be three. Now these two's are going to cancel, meaning we're gonna have one times three, which is just three, and that's going to be area one. So we've gone ahead and calculated this first shaded region right here, but now let's figure out what these other two areas are. Well, for area two, that's gonna be this region right here. And for area two, notice we have another triangle that forms. I know that the area of a triangle is one half base times height. Now what exactly is the base? Well, the base goes from this value of two all the way over to six. The difference between six and two would be four. So this has a base of four. Now notice our height goes from a y value of one all the way up to a y value of three. The difference between three and one is two. So that's gonna be that there. And then we cancel this one half of this two, giving us area of four. So an area of four is going to be this shaded region right here for area two, and that's going to be this next piece. Now lastly, we need to figure out what area three is down here. And for area three, we'll notice we're dealing with a rectangle. The area of a rectangle is base times height. Now it's clear that the base is gonna go from two all the way to six. The difference between six and two is four, and this is gonna be multiplied by the height. The height goes all the way from zero to one on the y axis. So that's gonna go to one, meaning we're going to have four as area three. So now we've gone ahead and figured out what these areas are. So these are the shaded regions here, and notice this gives us the entire area of this shape underneath. So we figured out that we had three, we had four, and then we had four again. Now what I'm ultimately trying to figure out in this problem is the integral that we're given up here. So trying to figure out what the integral is from zero to six of our function f of x with respect to x. But I've already figured out that the integral or the area under the curve of this function is going to be all of these numbers added together. And since the integrals from zero to six simply represents all of these numbers added together, the area from zero to six on this graph on the x axis, well, that means that adding these numbers will give us the solution to this integral. So we're going to have three plus four plus four. We know that three plus four is seven, and then we add four to that. We're going to get 11. So 11 would be the result and the area under this curve, and that is the solution to this problem. So that is how you can find definite integrals, which is just the area underneath the curve of a function based on these values right here on the x axis. So hope you found this video helpful. Let's go ahead and get some more practice and try some more problems. See you in the next one.

8. Definite Integrals

Introduction to Definite Integrals

Business Calculus

8. Definite Integrals

Introduction to Definite Integrals

Business Calculus 때문에 힘들어하고 계신가요?

저희를 신뢰하는 수천 명의 학생들이 시험 합류하세요!

객관식

Find $$\int$_0^6\!f$\left$(x$\right$)\,dx$ given the graph of $y=f$\left$(x$\right$)$ .

검증된 단계별 안내

Step 1: Understand that the integral ∫₀⁶ f(x) dx represents the area under the curve y = f(x) from x = 0 to x = 6. This area can be calculated by summing the areas of geometric shapes formed by the graph.

Step 2: Break the graph into distinct geometric shapes between x = 0 and x = 6. In this case, the graph forms a triangle from x = 0 to x = 2, another triangle from x = 2 to x = 4, and a trapezoid from x = 4 to x = 6.

Step 3: Calculate the area of the first triangle (x = 0 to x = 2). The base is 2 units (from x = 0 to x = 2), and the height is 3 units (y = 3 at x = 2). Use the formula for the area of a triangle: A = 1/2 × base × height.

Step 4: Calculate the area of the second triangle (x = 2 to x = 4). The base is 2 units (from x = 2 to x = 4), and the height is 3 units (y = 3 at x = 2). Use the same formula for the area of a triangle: A = 1/2 × base × height.

Step 5: Calculate the area of the trapezoid (x = 4 to x = 6). The bases are y = 3 (at x = 4) and y = 1 (at x = 6), and the height is 2 units (from x = 4 to x = 6). Use the formula for the area of a trapezoid: A = 1/2 × (base₁ + base₂) × height.

6:07m

다음 시청

마스터 Basic Rules for Definite Integrals가 Patrick Ford 짧은 영상으로 설명해 줍니다

Introduction to Definite Integrals 연습 세트

수석 튜터가 만든 문제 세트

전문가 영상 설명

Business Calculus 때문에 힘들어하고 계신가요?

Find ∫06f(x)dx\(\int\)_0^6\!f\(\left\)(x\(\right\))\,dx∫06f(x)dx given the graph of y=f(x)y=f\(\left\)(x\(\right\))y=f(x).

다음 시청

Introduction to Definite Integrals 연습 세트

Find $\(\int\)_0^6\!f\(\left\)(x\(\right\))\,dx$ given the graph of $y=f\(\left\)(x\(\right\))$ .