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# Finding definite integrals using area formulas

Since definite integrals are the net area between a curve and the x-axis, we can sometimes use geometric area formulas to find definite integrals. See how it's done.

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• So what if he combined the last two problems into one - sum1-6 f(x)dx? Is it just the sum of the areas of the triangle and semicircle, 6 + (-pi/2)? • I thought area under the curve meant that there couldn't be negative numbers • This is very convenient to us with or knowledge so far, but what would we do if the equation was shifted down 1 unit (or less)? How would we do the first problem, using our current knowledge, without using trapezoidal sums or Riemann sums, since we're just dealing with a semi-circle? • Interesting question - I think you could still solve it geometrically, it's just a significantly tougher problem. Check out this resource: https://www.mathopenref.com/segmentarea.html

You could compute the central angle given the radius of the circle and the height offset (however much you wanted to "shift the function down"). Computing the area of the circle segment that's greater than 0 gives the positive part of the area for x in `[-6, -2]`. The negative part could then be found by computing the area that's within the semicircle but outside of the positive segment, then subtracting that from the rectangle formed by `y=0` and `y=heightOffset` to get the negative area for x in `[-6, -2]`. (Disclaimer: I'm just sort of mulling through this in my head, so I'm not sure if that's exactly right, but it seems like the correct approach.)

That being said, the point of this problem is to be relatively convenient and provide a good way to grasp the concept at hand (integrals as areas under curves). The alternatives you listed are designed to solve more complex problems; not everything has to be done using the least available knowledge.
• Did sal forget to also calculate the other triangle in the interval [1,4]? .5(3*4)= 6 and .5(1*4)=2 then 2 + 6 = 8 I believe
(1 vote) • If we were to switch places with the numbers for each integral, for example, if we switched the 4 by the 6 and vice-versa in the 4th example shown in the video, would the area of the semi-circle be considered positive despite being below the x-axis? (I'm new to integration so I don't know much about all the properties yet, but I think I saw that somewhere, so I am not so sure if what I said is true) • Area is always positive.

However any area underneath the x-axis is negative when perform the integration. If you remember the explanation Sal gave using rectangles to approximate an area you realise if f(x)<0 then area of the rectangle will come out as negative f(x) dx is less than 0.

If switch the bounds of the integrand then the result will switch signs.

Try integrating from some function f(x) from a to b will lead result of

F(b)-F(a)

while swapping the bounds gets you

F (a)-F (b) = -( F (a) - F (b) ) which is opposite the above example