Question

11. ? 1/?(1-3x) dx Type 3 = ? (1-3x)^-1/3 dx -1/3 ((1-3x)^-1/3+1 / (-1/3 + 1)) + C = -1/2 (1-3x)^2/3 + C ? dx/?(1-3x) = -1/2(1-3x)^2/3 + C 12. ? x?(2x²+1) dx Type 1 ? x?(2x²+1) dx Substitute 2x²+1=t xdx = dt/4 ? x?(2x²+1) dx = 1/6(2x²+1)^3/2 + C ? 1/4 ?(t) dt 1/4 * 2/3 t^3/2 + C 13. ? (ln2x+1)³ 3/x dx Type 1 ? (ln2x+1)³ 3/x dx Substitute ln2x+1=t 1/2x * 2dx = dt dx/x = dt ? 3t³ dt 3t^4/4 + C ? (ln2x+1)³ 3/x dx = 3/4 (1+ln2x)^4 + C 14. Shade and find the area of the region between y=x²-1 and y=2x²+1 from x=0 to x=2. Area enclosed by curves in [0, 2] A = ? |f(x) - g(x)| dx = ? |x² - 1 - (2x² + 1)| dx ? |1 - x² - 2| dx = 20/3 sq units y=2x²+1 y=x²-1 (1,0) (2,0)

          11. ? 1/?(1-3x) dx Type 3
= ? (1-3x)^-1/3 dx
-1/3 ((1-3x)^-1/3+1 / (-1/3 + 1)) + C
= -1/2 (1-3x)^2/3 + C
? dx/?(1-3x) = -1/2(1-3x)^2/3 + C

12. ? x?(2x²+1) dx Type 1
? x?(2x²+1) dx
Substitute 2x²+1=t
xdx = dt/4
? x?(2x²+1) dx = 1/6(2x²+1)^3/2 + C
? 1/4 ?(t) dt
1/4 * 2/3 t^3/2 + C

13. ? (ln2x+1)³ 3/x dx Type 1
? (ln2x+1)³ 3/x dx
Substitute ln2x+1=t
1/2x * 2dx = dt
dx/x = dt
? 3t³ dt
3t^4/4 + C
? (ln2x+1)³ 3/x dx = 3/4 (1+ln2x)^4 + C

14. Shade and find the area of the region between y=x²-1 and y=2x²+1 from x=0 to x=2.
Area enclosed by curves in [0, 2]
A = ? |f(x) - g(x)| dx
= ? |x² - 1 - (2x² + 1)| dx
? |1 - x² - 2| dx = 20/3 sq units
y=2x²+1
y=x²-1
(1,0) (2,0)

11. ? 1/?(1-3x) dx Type 3
= ? (1-3x)^-1/3 dx
-1/3 ((1-3x)^-1/3+1 / (-1/3 + 1)) + C
= -1/2 (1-3x)^2/3 + C
? dx/?(1-3x) = -1/2(1-3x)^2/3 + C

12. ? x?(2x²+1) dx Type 1
? x?(2x²+1) dx
Substitute 2x²+1=t
xdx = dt/4
? x?(2x²+1) dx = 1/6(2x²+1)^3/2 + C
? 1/4 ?(t) dt
1/4 * 2/3 t^3/2 + C

13. ? (ln2x+1)³ 3/x dx Type 1
? (ln2x+1)³ 3/x dx
Substitute ln2x+1=t
1/2x * 2dx = dt
dx/x = dt
? 3t³ dt
3t^4/4 + C
? (ln2x+1)³ 3/x dx = 3/4 (1+ln2x)^4 + C

14. Shade and find the area of the region between y=x²-1 and y=2x²+1 from x=0 to x=2.
Area enclosed by curves in [0, 2]
A = ? |f(x) - g(x)| dx
= ? |x² - 1 - (2x² + 1)| dx
? |1 - x² - 2| dx = 20/3 sq units
y=2x²+1
y=x²-1
(1,0) (2,0)

Added by Rebecca N.

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