Calculus Problems: Integrals, Derivatives, and Area Calculation
Let f(x) = -x2 + 2x + 3. The graph of f is shown in the following diagram.
(a) Find ∫(-x2 + 2x + 3) dx
∫-x2 dx + ∫2x dx + ∫3 dx
-x(2+1)/(2+1) + 2x(1+1)/(1+1) + 3x + C
-x3/3 + 2x2/2 + 3x + C
= -x3/3 + x2 + 3x + C
(b) Find the area of the shaded region.
-(2)3/3 + (2)2 + 3(2) – [-(1)3/3 + (1)2 + 3(1)]
-8/3 + 4 + 6 – (-1/3 + 1 + 3)
-8/3 + 10 – (-1/3 + 4)
-8/3 + 10 = (-8 + 30)/3 = 22/3
-1/3 + 4 = (-1 + 12)/3 = 11/3
22/3 – 11/3 = 11/3 u2
Answer: 11/3 = 3.6 u2
2. Let f'(x) = 3x2 – 3. Given that f(2) = 1, find f(x).
∫(3x2 – 3) dx = ∫3x2 dx – ∫3 dx
3x(2+1)/(2+1) – 3x + C
3x3/3 – 3x + C
= x3 – 3x + C
f(x) = x3 – 3x + C
f(2) = (2)3 – 3(2) + C = 1
8 – 6 + C = 1
2 + C = 1
C = 1 – 2
C = -1
Answer: x3 – 3x – 1
3. Let f'(x) = 6x2 + 2x – 1. Given that f(2) = 5, find f(x).
∫(6x2 + 2x – 1) dx = ∫6x2 dx + ∫2x dx – ∫1 dx
6x3/3 + 2x2/2 – x + C
= 2x3 + x2 – x + C
f(2) = 5
2(2)3 + (2)2 – (2) + C = 5
16 + 4 – 2 + C = 5
18 + C = 5
C = 5 – 18
C = -13
Answer: 2x3 + x2 – x – 13
4. Let f(x) = 3x2 – 4x. The graph of f is shown in the following diagram.
(a) Find ∫(3x2 – 4x) dx
∫3x2 dx – ∫4x dx
3x3/3 – 4x2/2 + C
= x3 – 2x2 + C
(b) Find the area of the region enclosed by the graph of f, the x-axis, and the lines x = 2 and x = 4.
[(4)3 – 2(4)2] – [(2)3 – 2(2)2]
64 – 32 – (8 – 8)
= 32
Answer: 32 u2
5.
∫15 2f(x) dx = ∫15 f(x) dx + ∫15 3 dx
2F(x) |15 = F(x) |15 + 3x |15
2[F(5) – F(1)] = [F(5) + 3(5)] – [F(1) + 3(1)]
2(F(5) – F(1)) = F(5) + 15 – F(1) – 3
2(6) = 6 + 12
Answer: 12, Answer: 18
∫08 (1/(2x – 1)) dx
u = 2x – 1
du = 2 dx
dx = du/2
∫08 (1/u) (du/2) = (1/2) ∫08 (1/u) du
= (1/2) [ln|u|] |08 + C
= (1/2) [ln|2x – 1|] |08 + C
(1/2) [ln|2(8) – 1| – ln|2(0) – 1|] + C = (1/2) [ln(15) – ln(1)] + C
(1/2) ln(15) + C
(1/2) ln|2x – 1| + C = 5
(1/2) ln|2(1) – 1| + C = 5
(1/2) ln(1) + C = 5
C = 5
Result: (1/2) ln|2x – 1| + 5
7. The graph of a function f passes through the point (ln 2, 15). Given that f'(x) = 9e3x, find f(x).
∫ 9e3x dx
9 ∫ e3x dx
u = 3x
du = 3 dx
dx = du/3
9 ∫ eu (du/3) = 9/3 ∫ eu du
= 3eu + C
= 3e3x + C
3e3(ln 2) + C = 15
3eln(23) + C = 15
3(8) + C = 15
24 + C = 15
C = 15 – 24
C = -9
Result: 3e3x – 9
8. The graph of f passes through the point (π/12, 2). Given that f'(x) = 2cos(2x), find f(x).
∫ 2cos(2x) dx
2 ∫ cos(2x) dx
u = 2x
du = 2 dx
dx = du/2
2 ∫ cos(u) (du/2) = ∫ cos(u) du
= sin(u) + C
= sin(2x) + C
sin(2(π/12)) + C = 2
sin(π/6) + C = 2
1/2 + C = 2
C = 2 – 1/2
C = 3/2
Result: sin(2x) + 3/2
10.
∫0k (1/(1 + 2x)) dx = ln 3
u = 1 + 2x
du = 2 dx
dx = du/2
(1/2) ∫0k (1/u) du = (1/2) [ln|u|] |0k
= (1/2) [ln|1 + 2x|] |0k
(1/2) [ln|1 + 2k| – ln|1 + 2(0)|] = ln 3
(1/2) [ln(1 + 2k) – ln(1)] = ln 3
(1/2) ln(1 + 2k) = ln 3
ln(1 + 2k) = 2 ln 3
ln(1 + 2k) = ln(32)
1 + 2k = 9
2k = 8
k = 4
Result: k = 4
This is exercise 9
11.
