Help checking my work please!

Question

The region R is a rectangle with vertices P (a,ln a), Q (a, 0), S (3, 0), and T (3, ln a), where 1 < a < 3
(graph: gyazo.com/9e52cd3cd10fad297232efadb9f94d87)

A. Write an expression that gives the area of the rectangle as a function of a.
Area of a rectange = length x breadth. 1 < a < 3 so length = 3-a. The breadth = ln a.
So area, A = (3-a)•(ln a)

B. The area of the rectangle is maximized for some c between 1 and 3. Write the expression you would need to solve in order to find c. You don’t need to find c.
3/c - 1 - ln(c) = 0 such that -3/(c^2) - 1/c < 0.

C. What is the rate of change of the area when a =3, if a is decreasing at a rate of da/dt = − 0.5 units / sec? You don’t need to simplify, of course. Use appropriate units.
A = (3 - a)•ln(a)
dA/dt = -ln(a)(da/dt) + (3 - a)(da/dt)/a
= -ln(3)(-.5)un/sec + (3 - 3)(-.5)un/(3sec)
= .5•ln(3)un/sec
= ln(√3) units/sec

D. Again, a is decreasing at the rate of da/dt = − 0.5 units / sec. At a = 2.9, is the area of the rectangle increasing or decreasing? How do you know? Use appropriate units in your answer.
dA/dt = -ln(a)(da/dt) + (3 - a)(da/dt)/a
= -ln(2.9)(-.5)un/sec + (3 - 2.9)(-.5)un/(2.9sec)
= .5•ln(2.9)un/sec - .05un/(2.9sec)
≈ 0.515units/sec
> 0 units/sec
Hence, the area of the rectangle is increasing at a = 2.9

2. The shaded regions R1 and R2, shown below, are enclosed by the graphs of f(x)= −x^2 and g(x)= −2^x. (gyazo.com/7bcf02392ff69e2a1280588308342e8e)

A. Find the x- and y-coordinates of the three points of intersection of the graphs of f and g
I need help on this one!

B. Estimate the area of R1 to three reasonable places past the decimal.
The area of R1 is the integral from x ~ -0.76667 to x = 2 of
(-x^2+2^x) dx. Integrating these functions gives
-(1/3)x^3 + 2^x/(ln 2)
At the upper limit, the value is -8/3 + 4 ln(2)
At the lower limit, the value is about 0.55762
So area R1 is about -2.66667 + 2.77250 - 0.55762 = about 0.4517

C. Find the EXACT area of R2. (No calculator approximations; don’t simplify.)
The area of R2 is the integral from x = 2 to 4 of
(x^2-2^x) dx. Integrating these functions gives
(1/3)x^3 - 2^x/(ln 2).
area of R2 is exactly (64/3) - 16 ln(2) - (8/3) + 4 ln(2)
(56/3) - 12 ln(2)

D. Find the exact volume of the solid formed by revolving R2 about the x-axis. (No calculator approximations; don’t simplify.)
Area of each annulus = pi x^4 - pi 2^(2x)
Thickness of each annulus = dx
Integrate A dx from x = 2 to x = 4
The integrated function is (pi/5) x^5 - (pi/ln 2) [ 2^(2x) ]
The volume is (1024 pi/5) - (256 pi/ln 2) - (32 pi/5) + (16 pi/ln 2).
pi [ 992/5 - (240 pi/ln 2) ]

Answer 1

#1 looks ok
#2 looks ok
good work

Answer 2

Do you think you could help me on 2A?

Answer 3

Hey! I think you made a mistake in 2B, remember that the derivative is (2^x/ln2), you multiplied by ln 2.
Also, for 2A, graph the two functions and write down the three coordinates where the two graphs meet (-0.7667,-0.588), (2,-4), and (4,-16)

Answer 4

I think you're good