Question

Find the bounded area enclosed by the graphs of $f$ and $g$ when $f(x) = 3x - x^2 + 2$, $g(x) = x + 2$. 1. Area = $\frac{7}{3}$ sq.units 2. Area = $\frac{1}{3}$ sq.units 3. Area = $\frac{5}{6}$ sq.units 4. Area = $\frac{4}{3}$ sq.units 5. Area = $\frac{11}{6}$ sq.units

          Find the bounded area enclosed by the
graphs of $f$ and $g$ when
$f(x) = 3x - x^2 + 2$, $g(x) = x + 2$.
1. Area = $\frac{7}{3}$ sq.units
2. Area = $\frac{1}{3}$ sq.units
3. Area = $\frac{5}{6}$ sq.units
4. Area = $\frac{4}{3}$ sq.units
5. Area = $\frac{11}{6}$ sq.units
        
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Find the bounded area enclosed by the
graphs of f and g when
f(x) = 3x - x^2 + 2, g(x) = x + 2.
1. Area = (7)/(3) sq.units
2. Area = (1)/(3) sq.units
3. Area = (5)/(6) sq.units
4. Area = (4)/(3) sq.units
5. Area = (11)/(6) sq.units

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the bounded area enclosed by the graphs of f and g when f(x)=3x-x^(2)+2,g(x)=x+2. Area =(7)/(3) sq.units Area =(1)/(3) sq.units Area =(5)/(6) sq.units Area =(4)/(3) sq.units Area =(11)/(6) sq.units Find the bounded area enclosed by the graphs of f and g when O 1. Area 7 f(x)=3x-x2+2,g(x)=x+2 1 O 2. Area = sq.units 5 16 sq.units 3.Area 4 4.Area= sq.units 11 - sq.units 6 5.Area
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Transcript

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00:02 In this question, two functions are given.
00:03 First is f of x, which is 10 over square root x squared plus 1, and second one is g of x, which is x square plus 3x.
00:12 We need to calculate the area between these graphs of these two functions.
00:18 And we can use the graphing calculator to determine the points of intersection.
00:25 So first of all, we will plot the graphs of these two functions and determine the points of intersection using the graphing calculator.
00:33 So we plotted the graphs of these two functions.
00:36 The red one curve is for f of x and the blue is for g of x.
00:41 Now we also shaded the region between these two curves and by using the graphing calculator we got the points of intersection which are the first point of intersection is negative 3 .704, 2 .607 and second one is 1 .359.
01:00 5 .926.
01:02 Now we will use only up to two decimal places as given in question.
01:09 So we can see the lower limit for the integral will be as a which will be equal to minus 3 .70 and the upper limit will be b which will be x equals to 1 .35.
01:24 So the upper limit will be b will be 1 .35...
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