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Assignment 2

1. The area of the region bounded by the curves f(y) = y^2 + 2 and g(y) = 6 is 32/3. 2. The area of the region bounded by the graph of f(x) = 1/(1+x^2) and its tangent line at x=1 is (π-3)/4. 3. The derivative of y = cosh(sin x) − xesinh x is sinh(sin x) cos x − esinh x − xesinh x cosh x.

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Vinícius Castellani
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108 views2 pages

Assignment 2

1. The area of the region bounded by the curves f(y) = y^2 + 2 and g(y) = 6 is 32/3. 2. The area of the region bounded by the graph of f(x) = 1/(1+x^2) and its tangent line at x=1 is (π-3)/4. 3. The derivative of y = cosh(sin x) − xesinh x is sinh(sin x) cos x − esinh x − xesinh x cosh x.

Uploaded by

Vinícius Castellani
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Calculus 2 with The Math Sorcerer

1. Find the area of the region bounded by the curves f (y) = y 2 + 2 and g(y) = 6.

1
2. Find the area of the region bounded by the graph of f (x) = and the tangent line to
  1 + x2
1
the graph of f (x) at 1, .
2

3. Differentiate y = cosh(sin x) − xesinh x .

Z
sinh x
4. Integrate dx.
1 + sinh2 x

5. Find the equation of the tangent line to the graph of h(x) = xcosh x at (1, 1).
SOLUTIONS Z 2
32
1. The area is A = (6 − (y 2 + 2)) dy = . If you look at your graphed region you may
−2 3
Z 2
32
notice it is easier to compute A = 2 (6 − (y 2 + 2)) dy = . Another way to see this shortcut
0 3
2
is to notice that the integrand h(y) = 4 − y is an even function of y. Recall that for an even
Z a Z a
function f (x) we have f (x) dx = 2 f (x) dx.
−a 0

1
2. To graph the function f (x) = you can use calculus I techniques or your calculator.
1 + x2
Next you need to graph the tangent line to the graph of f (x) at x = 1. The slope of the tangent
1 1
line is f ′ (1) = − so using the equation of a line yields y = − x + 1. The desired area is then
Z 1 2  2
1 1 π−3
− − x+1 dx =
0 1 + x2 2 4

3. y ′ = sinh(sin x) cos x − esinh x − xesinh x cosh x.

1
4. Use cosh2 x − sinh2 x = 1. The solution is − + C.
cosh x

5. The easiest way to do this problem is to let y = xcosh x , so ln y = cosh x ln x. Now use
implicit differentiation, then plug in x = 1 or solve for y ′ and then plug in x = 1, the choice is
yours. The slope of the tangent line is then y ′ (1) = cosh 1. The equation of the tangent line is
y = cosh(1)x+1−cosh 1. Note cosh 1 ≈ 1.543, so you could write the answer as y = 1.543x−.543
as well.

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