# Simple integration (high school)

• synkk
In summary, when finding the integral between two limits, it is essentially finding the signed area between the graph of the function and the x-axis. The integral of a function from a to b is the area above the x-axis, while the integral of a negative function from a to b is the negative of the area above the x-axis. When subtracting two integrals, the result is the total area above the x-axis, minus the area above the curve where it is negative. This can be illustrated using graphs, as shown in the conversation. Some math programs, such as Wolfram Alpha, can generate graphs to help visualize these concepts.
synkk

For part c)

The equation of ON is y = -0.5x which is correct;
i know to solve it i need to find the integral between the limits 5/2 and 0 of the area of the curve - the area of the line, but what i don't understand is why that is.

If I take the integral of the limits 2 and 0 of the curve, which area I'm i actually finding? It seems like I am doing things without understanding them, could someone show me a sketch? Also, if I'm finding the integral of the line y = -0.5x of the limits 5/2 and 0, which area am i exactly finding? Why is it the integral between the two equations which is the shaded bit?

If anyone could clear this up would be great.

If you integrate f(x) from a to b, you get the (signed) area of the region between the graph of y= f(x) and the x-axis. The integral of $2x- x^2$ with respect to x from x=0 to x= 2 is the area below the curve and above the x-axis. The integral of -0.5x with respect to x from x= 0 to x= 2.5 is -1 times the area between the graph of y= -0.5 x and the x-axis.

The graph of $y= 2x-x^2$ goes below the x-axis so the signed area between it and the x-axis is negative- but not "as negative" as -0.5x so the integral of $(2x- x^2)- (-0.5x)$ is still positive.

HallsofIvy said:
If you integrate f(x) from a to b, you get the (signed) area of the region between the graph of y= f(x) and the x-axis. The integral of $2x- x^2$ with respect to x from x=0 to x= 2 is the area below the curve and above the x-axis. The integral of -0.5x with respect to x from x= 0 to x= 2.5 is -1 times the area between the graph of y= -0.5 x and the x-axis.

The graph of $y= 2x-x^2$ goes below the x-axis so the signed area between it and the x-axis is negative- but not "as negative" as -0.5x so the integral of $(2x- x^2)- (-0.5x)$ is still positive.

Is the area always from the graph, to the x axis? Also is this what you mean of the area (the "A" bit):

If so, I still don't understand how subtraction the two would give you the area needed.

The area you get from integrating like this is always measured from the x-axis (if you integrate with respect to x that is). That's how the subtraction thing works. The total area under the line ON (well above actually), minus the total area above the curve from where the curve is negative to the intersection point.

I made this image for you, it should explain it. Just remember that the area under the x-axis will be negative area!

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QuarkCharmer said:
The area you get from integrating like this is always measured from the x-axis (if you integrate with respect to x that is). That's how the subtraction thing works. The total area under the line ON (well above actually), minus the total area above the curve from where the curve is negative to the intersection point.

I made this image for you, it should explain it. Just remember that the area under the x-axis will be negative area!

So wouldn't the area we are finding by subtracting the two be this area?

synkk said:
So wouldn't the area we are finding by subtracting the two be this area?

Not quite. Figure 1 is $\displaystyle\int_0^{\frac{5}{2}} (2x-x^2) \, dx$ and Figure 2 is $\displaystyle\int_0^{\frac{5}{2}} -\frac{1}{2}x \, dx$. Do you now see why it's correct to subtract the integrals?

scurty said:
Not quite. Figure 1 is $\displaystyle\int_0^{\frac{5}{2}} (2x-x^2) \, dx$ and Figure 2 is $\displaystyle\int_0^{\frac{5}{2}} -\frac{1}{2}x \, dx$. Do you now see why it's correct to subtract the integrals?

I do now, thank you very much, can I ask what program that is you're using for the images?

Glad to hear it! I used the Wolfram Alpha website to generate the graphs of the functions. I the. Saved the image to my computer, went into Microsoft Paint, drew the vertical line at 2.5 myself and filled it in myself! :)

I'm sure there are math programs out there that can do better but that's what I did.

scurty said:
Glad to hear it! I used the Wolfram Alpha website to generate the graphs of the functions. I the. Saved the image to my computer, went into Microsoft Paint, drew the vertical line at 2.5 myself and filled it in myself! :)

I'm sure there are math programs out there that can do better but that's what I did.

Okay, if anyone knows any please do let me know, thanks.

scurty said:
Glad to hear it! I used the Wolfram Alpha website to generate the graphs of the functions. I the. Saved the image to my computer, went into Microsoft Paint, drew the vertical line at 2.5 myself and filled it in myself! :)

I'm sure there are math programs out there that can do better but that's what I did.

Okay, if anyone knows any please do let me know, thanks.

## 1. What is simple integration?

Simple integration, also known as indefinite integration, is a mathematical process used to find the antiderivative of a function. It involves reversing the process of differentiation and finding the original function from its derivative.

## 2. How is simple integration different from complex integration?

Simple integration only involves finding the antiderivative of a basic function, while complex integration involves finding the antiderivative of more complex functions using techniques such as substitution, integration by parts, or partial fractions.

## 3. What is the purpose of simple integration?

The purpose of simple integration is to find the original function from its derivative, which is useful in various fields of science and engineering, such as physics and economics. It also helps in solving problems involving area, volume, and motion.

## 4. Can you provide an example of simple integration?

One example of simple integration is finding the antiderivative of the function f(x) = 3x^2. Using the power rule, we know that the antiderivative of x^n is (1/(n+1))x^(n+1). Therefore, the antiderivative of f(x) is (1/3)x^3 + C, where C is the constant of integration.

## 5. What are some common techniques used in simple integration?

Some common techniques used in simple integration include the power rule, u-substitution, and factoring. These techniques are used to simplify the function and make it easier to find the antiderivative.

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