MATH 16 Calculus Improper Integrals & Density Function Questions

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MATH 16

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Math 16 B: Practice Final 1. You have an account with interest compounded continuously with 2% annual interest rate. After 2 years you have 2000 $. How much money did you have in the beginning? 2. The amount of bacteria in a petri dish is currently 100 and is expected to be 20000 in 2 days. Assuming that the bacteria amount can be described via exponential growth, how many bacteria will be there in 1 day? 3. Decide, with justification, if Z (1 + x cos x)esin x dx = xe1+sin x + C 4. Calculate the midpoint approximation M3 to the integral Z 7/2 1/2 1 √ dx 1+ x and calculate the exact value of the integral as well. 5. Draw a sketch of the region bounded by the graphs of y = (x − 1)2 − 2 and y = x − 3. Compute the area of the region. 6. Calculate the following indefinite integrals: (a) Z 3x2 e−5x + ln x dx x (b) Z 2 2 (ex + sin(x4 )) · (xex + 2x3 cos(x4 )) dx (c) Z x − 10 dx x + 10 7. Calculate the following indefinite integrals: (a) Z (1 + sin x)4 cos x dx (b) Z sec x tan x − sin x dx (c) Z 2 sec2 x + 3 tan2 x dx 1 8. Calculate the following improper integrals: (a) −1 Z −∞ 1 dx x2 (b) 2 Z 1 1 dx x2 − 4 Z ∞ (c) 0 1 dx x2 9. Suppose Z 1 2f (x)f 0 (x) dx = 10 0 and f (0) = 0. Find the value of f (1)2 . 10. Find all values (if any) of c ≥ 0 such that 3x2 2 is a probability density function on [−c, c]. For such a value of c calculate the mean µ, the median m, and the variance V . f (x) = 2
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1

1. Denote the initial amount as 𝑥, then 𝑥 ∙ 𝑒 2∙2% = 𝑥𝑒 25 = 2000.
1

Therefore, 𝑥 = 2000𝑒 −25 ≈ 𝟏𝟗𝟐𝟏. 𝟓𝟖 ($).

2. In 𝑛 days, there will be 100𝑒 𝑎𝑛 bacteria for some 𝑎 such that 100𝑒 2𝑎 = 20000.
We see 𝑒 2𝑎 = 200, 𝑒 𝑎 = √200 = 10√2. Thus in 1 day there will be
100𝑒 𝑎 = 1000√2 ≈ 𝟏𝟒𝟏𝟒 bacteria.

3. To check, differentiate both sides. The left will give the function under integral, i.e.
(1 + 𝑥 cos 𝑥)𝑒 sin 𝑥 , the right will give
′
(𝑥𝑒 1+sin 𝑥 ) = 𝑒 1+sin 𝑥 + 𝑥𝑒 1+sin 𝑥 cos 𝑥 = 𝑒 1+sin 𝑥 (1 + 𝑥 cos 𝑥).
We see these functions are not the same (the second is the first multiplied by 𝑒).

1 7

1 3

3 5

5 7

4. To find 𝑀3 , divide the segment [2 , 2] by three equal segments [2 , 2], [2 , 2], [2 , 2] of
length 1 each and take their midpoints 1, 2, 3. The function values are
1
1
1
𝑓(1) = , 𝑓(2) =
, 𝑓(3) =
.
2
1 + √3
1 + √2
𝟏

𝟏

𝟏

This way 𝑀3 = 1 ∙ (𝑓(1) + 𝑓(2) + 𝑓(3)) = 𝟐 + 𝟏+√𝟐 + 𝟏+√𝟑 ≈ 𝟏. 𝟐𝟖𝟎.
The exact value of this integral is
∫

7
2

1
2

𝑑𝑥

1
7
𝑑𝑥
= |𝑢 = 1 + √𝑥, 𝑢 ∈ [1 + √ , 1 + √ ] , 𝑑𝑢 =
, 𝑑𝑥 = 2√𝑥𝑑𝑢 = 2(𝑢 − 1)𝑑𝑢| =
2
2
1 + √𝑥
2√𝑥

7
7
1 + √2
1+√
2
7
1
=∫
(2 − ) 𝑑𝑢 = (2𝑢 − 2 ln 𝑢) 2 = 2 (√ − √ ) − 2 ln
=
1
𝑢
2
2
√1
1
1+√
𝑢=1+
2
2
1 + √2
√𝟐 + √𝟕
= (√𝟏𝟒 − √𝟐) − 𝟐 𝐥𝐧
≈ 𝟏. 𝟐𝟖𝟖.
√𝟐 + 𝟏
7
1+√
2

5. The picture:

As we see (and can solve), the region is between 𝑥 = 1 and 𝑥 = 2 and on this segment
𝑥 − 3 ≥ (𝑥 − 1)2 − 2, thus the area is
2

2
2

∫ (𝑥 − 3 − ((𝑥 − 1) − 2))𝑑𝑥...