Question

The graph of f is shown with the area of the region bounded by the graph and the x-axis on [0,8]. Find the values of the definite integrals a) \int_0^3 f(t) dt b) \int_0^8 f(t) dt c) \int_0^8 |f(t)| dt

          The graph of f is shown with the area of the region bounded by the graph and the x-axis on [0,8].
Find the values of the definite integrals
a) \int_0^3 f(t) dt
b) \int_0^8 f(t) dt
c) \int_0^8 |f(t)| dt
        
The graph of f is shown with the area of the region bounded by the graph and the x-axis on [0,8].
Find the values of the definite integrals
a) ∫0^3 f(t) dt
b) ∫0^8 f(t) dt
c) ∫0^8 |f(t)| dt

Added by Hector D.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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The graph of f is shown with the area of the region bounded by the graph and the x-axis on [0,8]. Find the values of the definite integrals: a) ∫f(t)dt b) ∫f(t)dt
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Transcript

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00:01 Alright, so we want to find, we have this area function, the fundamental theorem of calculus, and we can find these areas by looking at the graph.
00:09 So the first one, a of 2, i'm going from 0 to 2, it's this little region right here, which we're told is a quarter of a circle with a radius equal to 2.
00:20 Area of a circle is pi times the radius squared.
00:24 We only have a fourth of it, so we divide by 4, so this comes out to be pi.
00:29 Now, it is below the x -axis, so it is negative, so don't forget to make it negative because it's below the x -axis.
00:38 Alright, now, when we go all the way from 0 to 5, sorry about that, i sneezed, i didn't want to disrupt the video.
00:47 Alright, when i go from 0 to 5, that's going to take me all the way to right here, there's 5 right there.
00:55 Alright, well i just got to find the area of this red triangle, but don't forget, remember this part is part of what our formula is too, our a of x formula is, so i got to add the negative pi or subtract pi off.
01:12 So the area of that triangle, so the base is 3, the height is also 3, so base times height divided by 2, that comes out to be 4 .5...
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