Math 103. Homework 11. Solutions

1. Find the indefinite integral

ó õ (x1/3 -Öx +4) dx

Solution.

ó õ (x1/3 -x1/2 +4) dx = 1 1/3+1 x1/3 +1 - 1 1/2+1 x1/2+1 + 4x +C = 3 4 x4/3 - 2 3 x3/2 + 4x +C

2. Find the indefinite integral

ó õ x-1 (x2-2x+5)1/2 dx

Solution. The substitution

u = x2 -2x +5

du = ( 2x-2) dx = 2(x-1) dx

gives

ó õ x-1 (x2-2x+5)1/2 dx = 1 u1/2 1 2 du = u1/2+ C = ( x2 -2x +5)1/2 +C

3. Find the indefinite integral

ó õ (lnx)5 x dx

Solution. Substitute

u=lnx

du = 1 x dx

so that

ó õ (lnx)5 x dx = ó õ u5 du = 1 6 u6 + C = 1 6 (lnx)6 +C

4. [56, p. 464] The number of viewers of a weekly TV news-magazine show, introduced in the 1995 season, has been increasing at a rate of

3 æ ç è 2+ 1 2 t ö ÷ ø -1/3          (1 £ t £ 6)

million viewers/year in its t th year on the air. The number of viewers of the program during its first year on the air was 9(5/2)^2/3 million. Find how many viewers were expected in the 2000 season.

Solution. Call N(t) the number of viewers in the t-year. Then

N¢(t) = 3 æ ç è 2+ 1 2 t ö ÷ ø -1/3

Therefore,

N(t) = ó õ N¢(t) dt = ó õ 3 æ ç è 2+ 1 2 t ö ÷ ø -1/3   dt

To calculate this indefinite integral, do a change of variable

u = 2+ 1 2 t

du = 1 2 dt

so that the integral is

ó õ 3 u-1/3(2) du = 6 1 1-1/3 u1-1/3+C = 9 u2/3 +C

Substituting back to the variable t, we obtain

N(t) = 9 æ ç è 2+ 1 2 t ö ÷ ø 2/3   +C

To find out what is the value of the constant C, we note that

N(1) = 9 æ ç è 2+ 1 2 ö ÷ ø 2/3   +C = 9(5/2)2/3

so that C=0.

Therefore, in the 2000 season (when t=6), the expected number of viewers is

N(6) = 9 æ ç è 2+ 1 2 6 ö ÷ ø 2/3   » 26.3162

or 26,316,200 viewers.

5. Find the area of the region under the curve y=e^2x from x=0 to x=2.

Solution. The area is the definite integral

ó õ 2 0  e2x dx = 1 2 e2x ê ê ê 2 0  = 1 2 e4 - 1 2 e0 = 1 2 (e4-1)