B Learning Integrals with the Graph of y=x^2

[Stephanus]

Dear PF Forum,
I'd like to study integral.
But I just realize that I'm lack of basic integral.
Can I ask here?
Let y=x²
Here is the graph.

So,
1. dx is the distance between the red vertical lines? But it's very, very, very small distance.
2. f(x) * dx is the yellow area?
3. $\int_0^2 x^2\,dx$ is the green (plus yellow) area?
4. And one more thing. The blue line function can be derived from the derivative of f(x)?
For example the derivative of x² is 2x. So a line that crosses (x,f(x)) where x=1 is 2.
The blue line function is y = 2x.
If it crosses (x,f(x)) where x = 2, then the blue line function is y = 4x?
Thank you very much

 

[Mark44]

Dear PF Forum,
I'd like to study integral.
But I just realize that I'm lack of basic integral.
Can I ask here?
Let y=x²
Here is the graph.
View attachment 108202
So,
1. dx is the distance between the red vertical lines? But it's very, very, very small distance.

We usually call this $\Delta x$.

2. f(x) * dx is the yellow area?

If $x_i$ is a number in the subinterval at the base of the yellow region, then $f(x_i)\Delta x$ is the area of a rectangle that is approximately equal to the area of the yellow region. The smaller the value of $\Delta x$, the better the approximation is. Here we are approximating a roughly trapezoidal shape by one with a rectangular shape.

3. $\int_0^2 x^2\,dx$ is the green (plus yellow) area?

Yes

4. And one more thing. The blue line function can be derived from the derivative of f(x)?
For example the derivative of x² is 2x. So a line that crosses (x,f(x)) where x=1 is 2.

The blue line doesn't "cross" the curve -- it touches it at the point of tangency.

The blue line function is y = 2x.
If it crosses (x,f(x)) where x = 2, then the blue line function is y = 4x?

No. The slope of the tangent line (what you're calling the blue line) is 4, but if you draw this line you'll see that the tangent line intersects the y-axis below the origin. The line whose slope is 4, that is tangent to the graph of y = x² at (2, 4), has this equation: y = 4x - 4.

 

[FactChecker]

EDIT: Sorry. I took a break before submitting and this is mostly a repeat of @Mark44's answer above.

1. dx is the distance between the red vertical lines? But it's very, very, very small distance.

Yes.

2. f(x) * dx is the yellow area?

Yes. But there is a small range of x values in that short dx, so you pick one, say x₀, in that range and use f(x₀₎) * dx as the approximation of the yellow area. As dx gets smaller, the approximation gets better.

3. $\int_0^2 x^2\,dx$ is the green (plus yellow) area?

Yes.

4. And one more thing. The blue line function can be derived from the derivative of f(x)?

The derivative f(x₀) only gives you the slope of the line tangent at x₀. So figuring out the constant term of the line is still required.

For example the derivative of x² is 2x. So a line that crosses (x,f(x)) where x=1 is 2.

Except for the constant. In general, the equation of the tangent line is f'(x₀) * (x-x₀) + f(x₀)

 

[Stephanus]

We usually call this $\Delta x$.

Yes, I think it's just semantic here. I usually see f(u) du or F(v) dv.
But delta is more appropriate I think. The difference, right?
Thanks.

If $x_i$ is a number in the subinterval at the base of the yellow region, then $f(x_i)\Delta x$ is the area of a rectangle that is approximately equal to the area of the yellow region. The smaller the value of $\Delta x$, the better the approximation is. Here we are approximating a roughly trapezoidal shape by one with a rectangular shape.

Trapezoidal shape looks very similar to rectangle. Thanks.

3. $\int_0^2 x^2\,dx$ is the green (plus yellow) area?

Yes

Thanks

The blue line doesn't "cross" the curve -- it touches it at the point of tangency.

Yes, I should have drawn it more clearly. What I was going to draw is that the blue line did touch the curve. It touches not crosses. Thanks

No. The slope of the tangent line (what you're calling the blue line) is 4, but if you draw this line you'll see that the tangent line intersects the y-axis below the origin. The line whose slope is 4, that is tangent to the graph of y = x² at (2, 4), has this equation: y = 4x - 4.

Yes, I should have drawn it below the origin. Thanks. So it is slope? In high school we call it gradient.

Now I can at least read any reference about integration.
THanks.

 

[Stephanus]

Thanks @FactChecker . I suspect that all the answer should be yes. But I need confirmation. I'm going to read some calculus source. Thanks.

 

[Mark44]

Thanks. So it is slope? In high school we call it gradient.

Here the meanings are the same, but the term "gradient" is also used in functions of two or more variables, and in that context, the gradient of f (denoted $\nabla f$) is a vector that consists of the partial derivatives of f. For example, for a function f(x, y, z), $\nabla f$ is the vector $<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}>$ .

 

[FactChecker]

the gradient of f (denoted $\nabla f$) is a vector that consists of the partial derivatives of f.

I would only add that the gradient vector points in the direction of fastest increase of the function and its magnitude is the slope in that direction.