Solutions for Calculus Final Exam 1. See page 360 and 368 on the text book. 2. (a) See the solution of 3(c) of Midterm. (b) See the solution of 3(d) of Midterm. (c) y

Solutions for Calculus Final Exam 1. See page 360 and 368 on the text book.

2. (a) See the solution of 3(c) of Midterm.

(b) See the solution of 3(d) of Midterm.

(c) y³x² − yx + 2y²= x.

3y^{2 dy}_dxx²+ y³(2x) − ^dy_dxx − y + 4y^dy_dx = 1.

Thus dy

dx = 1 + y − 2y³x 3y²x²− x + 4y. (d) By definition

f^′(1) = lim

h→0

f (1 + h) − f(1) h

= lim

h→0

f (1 + h) − 1 h

=

( lim_h→0^(1+h)_h²⁻¹ = lim_h→0h + 2 = 2 if h = _n¹ lim_h→0^−(1+h)_h ²⁻¹ = lim_h→0^{−h2−2h−2}_h = −∞ if h 6= n¹

Thus f^′(1) does not exist.

(e)

dy

dx = d dx

Z x³−2x 1+x²

cos tdt

= cos(x³− 2x)(x³− 2x)^′ − cos(1 + x²)(1 + x²)^′

= (cos(x³− 2x))(3x²− 2) − (cos(1 + x²))(2x) 3. (a) Set x = tan u.

Then dx = sec²udu.

Z ₋₁

−√ 3

4

1 + x²dx =

Z _−π/4

−π/3

4

1 + tan²usec²udu

=

Z _−π/4

−π/3

4 1 + tan²u

1 cos²udu

=

Z _−π/4

−π/3

4

cos²u + sin²udu

=

Z _−π/4

−π/3

4du

= 4u|^−π/4_−π/3

= π/3 (b)

Z π/3 0

e^xsin xdx

(u = sin x, dv = e^xdx)

= sin xe^x|^π/30 − Z π/3

0

e^xcos xdx (u = cos x, dv = e^xdx)

= sin xe^x|^π/30 − cos xe^x|^π/30 − Z π/3

0

e^xsin xdx

= e^π/3(

√3 − 1

2 ) + 1 − Z π/3

0

e^xsin xdx Thus

Z π/3 0

e^xsin xdx = e^π/3(

√3 − 1 4 ) + 1

2 (c) R (x⁻³+ 3^−x)dx = −¹₂x⁻²−_{ln 3}¹ 3^−x+ C

(d) Let w =√

x + 1, dw = ₂^√1_xdx = _2(w−1)¹ dx Z

ln(√

x + 1)dx = 2 Z

(ln w)(w − 1)dw

(u = ln w, dv = (w − 1)dw)

= 2[(ln w)(1

2w²− w) − Z

(1

2w²− w)1 wdw]

= (ln w)(w²− 2w) − 2 Z

(1

2w − 1)dw

= (ln w)(w²− 2w) −1

2w²+ 2w + C

= ln(√

x + 1)[(√

x + 1)²− 2(√

x + 1)] −1 2(√

x + 1)²+ 2(√

x + 1) + C

= (x − 1) ln(√

x + 1) +√ x − 1

2x + C (e)

Z

cos²xdx =

Z 1 + cos 2x

2 dx

= x

2 +sin 2x 4 + C 4. (a) P = [x0, ..., x5] = [0,¹₅,²₅, ..., 1]

∆x_k= ¹₅ for k = 1, ..., 5

ck = x_k−1 are rational for k = 1, ..., 5 Then the Riemann Sum

5

X

k=1

f (ck)∆xk= 1 5(0 + 1

5 +2 5 +3

5 +4 5) = 2

5 (b) No

5. y =R _e−x

+e^x

2 dx = ¹₂(e^x− e^−x) + C And y(0) = C = 0.

Thus y = ¹₂(e^x− e^−x).

Page 2

6.

L(x) = Z

e^−0.1xdx

= −10e^−0.1x+ C L_∞ = lim

x→∞L(x) = lim

x→∞(−10e^−0.1x+ C) = C = 25 Thus L(x) = −10e^−0.1x+ 25

And L(0) = −10 + 25 = 15 7. (a)

f^′(x) = exp[3

2(1 − x

10)] − 3

20x exp[3

2(1 − x 10)]

f^′(x)







> 0 if x < ²⁰₃ increasing

= 0 if x = ²⁰₃

< 0 if x > ²⁰₃ decreasing f^′′(x) = exp[3

2(1 − x 10)]( 9

400x − 3 10) f^′′(x)







> 0 if x > ⁴⁰₃ concave up

= 0 if x = ⁴⁰₃ inflection point

< 0 if x < ⁴⁰₃ concave down Since f^′′(²⁰₃ ) < 0, f (x) has a local maximum at x = ²⁰₃. lim_x→∞f (x) = 0, asymptote: y = 0

(b) We solve x = f (x) = x exp[³₂(1 − ₁₀^x)]

⇒ x = 0, 10

(c) |f^′(0)| = exp(³₂) > 1, unstable

|f^′(10)| = | − ¹2| < 1, local stable (d) with oscillations.

Page 3