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Let $g(x)=\int_{0}^{x} f(t) d t,$ where $f$ is the function whose graph is shown.(a) Evaluate $g(x)$ for $x=0,1,2,3,4,5,$ and 6 .(b) Estimate $g(7)$(c) Where does $g$ have a maximum value? Where does it have a minimum value?(d) Sketch a rough graph of $g$.

$f(x)>0$$f(x)=0$$f^{\prime}(x)<0$

01:47

Frank L.

Calculus 1 / AB

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration

Ben M.

June 7, 2021

let g(x) = 4(x+1)^-2/3 and let f be the function defined by f(x)= d \int_{x}^{0} g(t) dt for x>0

let g(x) = 4(x+1)^-2/3 and let f be the function defined by f(x)= d \int_{x}^{0} g(t) dt for x>0 a) find f(26) b) determine the concavity of the graph of y= f(x) for x>0 . justify your answer c) let h be the function defined by h(x) = x-

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we are given a function F for given its graph. And we're told that G fx is the function which is the integral from zero to X. Of F F t D T. In part A were asked to evaluate the function G For values of X between zero and 6. Well, as G of X is defined, it's the area between the graph of F and T axis, with the areas below the axis being negative. And therefore looking at the graph of F. Well, first of all, G of zero, by definition is the integral from 0 to 0 of F F T. D. T. And since the limits on the integral are the same, this is zero. G of one is the integral from 0 to 1 of fft D. T. Looking at the graph of F. Well, The area under this part of the graph is two whole squares. And so this is simply too. Likewise, GF two is the integral from 0 to 2 of FTDT. Mhm Looking at the graphs, F Mhm. You see the integral from 0 to 1. We saw was four Plus. The integral from 1 to Which is .5 times two. Okay, for a total of five G of three is the integral from 0 to 3 of fft. So this is G F two and by the activity of integral G of two plus the integral from 2 to 3 of F F t D t. Yeah, From previous parts, this is five plus in the area of this triangle is .5 times for the height of the triangle For a total of seven. Yeah. Yeah. Finally, Do you have six Is the integral from 0 to 6 of FFTDT. Which from previous parts, this is G F three Plus the integral from 3 to 6 of the 50. And this is seven. And then this area well this is below the X axis. And so it's actually a negative area so seven minus. And this is a triangle into a square. So we have the area of the triangle is .5 times the height which is four plus the area of the square or rectangles. And say She has a height of two. Yeah. Which comes out to three. Then in part B Were asked to estimate GF seven. Well by activity of integral this is the same as Integral from 0 to 7 of Fft Which is equal to the integral from 0 to 6 of Fft Plus the integral from 6 to 7 of F 15. Using part A. This is Yeah. Mhm. Yeah. Yeah. Yeah. Three. Yeah. Mhm. Yeah. Okay. Plus say about one which is four. Then in part C. Were asked where our function G has a maximum value and where it has a minimum value mm Well we see that from X equals zero up to three. The area is all positive. So in fact the accumulative area G is greatest and therefore has a maximum value At x equals three. Mhm. Likewise we see that From x equals 3- seven. The area is negative. And so we see that are accumulative area G is leased or it has a minimum And x equals seven. Finally in part the rest to sketch a rough graph of the function G. Yeah. Yeah. Mhm. No. Before we do this notice that from part A we saw that G was in fact always increasing From zero up to three. Yeah. Mhm. And then when X equals three the function is at its maximum And then in the interval from 3 to 6 the function was decreasing. And so the graph of G Starts out at zero, Which is a maximum that x equals three and then decreases. Again, it looks something like this.

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