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Integrating scaled version of function

Sal uses a graph to explain why we can take a constant out of a definite integral.

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• I still don't intuitively understand the connection between definite integral and derivatives; area under the curve and rate of a function. I believe Sal has explained that somewhere, but is there one more specific video on that topic or should I just watch them all?
• at , Sal says "a positive c greater than zero". but shouldn't it be "greater than one" since this is a vertical stretch (not a compression, from the way it was drawn)? thanks!
• Sal said he chose to draw it with c = 3, just because he had to use some value. But what he showed will work for any value of c > 0. He says the line may scaled up or down depending on c.
• I understand the property, but when i think about geometrical scalar factors, in those problems, the factor would increase both dimensions.

Unless, when dealing those scalars, each dimension is only receiving half of the increase, as oppose to this where the one dimensions receives all. For example, if c were 2, a geometric shape would actually only increase in each dimension by 1.5.

Is this right?
• In geometry the scalar factor would apply to all the dimensions. But we are not dealing with an equation for a geometric figure here. We are applying the scalar to a function. Visually f(x) is just a wavy line. So there are not "both dimensions" to scale. The C is only affecting the f(x)/y-values by moving the line up or down. The integral is an operation telling you to find the area under that line between a and b. This is where you begin to see a shape because you draw in the boundaries (called limits of integration) to help you visualize the area underneath your line. But in actuality, the function you're working with is just a line and never connects to itself to encloses an area like a geometric shape would. If it did, it would crease to be a function (fails the vertical line test). Hope this helps.
• What if the function is dilated in the x direction?
e.g. how would we find the area from x=a to x=b of the function f(c*x) ?
• Good question!

The integral (or net area) from x=a to x=b of f(c*x) dx is the same as (1/c) times the integral of x=ca to x=cb of f(x) dx.

(This comes from using the u-substitution u=cx which gives du=c dx, expressing everything in terms of u and du, changing the limits of integration accordingly, and then simply replacing all the u’s by x’s in the end.)
• We have studied that the integral of x^n is x^n+1/n+1. Can you elaborate?
• Remember the power rule? The derivative of x^n is nx^(n-1). You are integrating x^n. Since you want the exponent to be n after you differentiate the integral, take a look at the power rule. The exponent will be one lower when you differentiate. So, you need to take it one up when you are integrating. That gives x^(n+1). However, that is not the answer. From the power rule, the exponent also "comes down" to be a coefficient. You want a coefficient of one after you differentiate the integral, so you need to put 1/(n+1) in front. Your final answer for the integral of x^n: (x^(n+1))/(n+1)+C.

Note: In your question, you forgot to put +C. That term is very important in indefinite integrals.
• I DO NOT understand how does doing integration help us to get the area under or below or whatever of a curve...........please someone help me understand the theory behind it....
• what if c is being added to the function (f(x)+c) but the limits stay the same?
• Then c(b-a) will be added to the area under f(x) to get the area of f(x)+c
• If integration gives the area under a curve, will it be always positive?
• When you multiply a function by a constant, you're just ''stretching'' it ''tightening'' it (above or below the x axis, depending on the sign of `c`).