Problem :
Evaluate
x(2x
^{2}+4)^{3}
dx
.
u | = 2x ^{2} +4 | ||
= 4x | |||
dx | = |
x u ^{3} | = u ^{3} du = | ||
= 2x ^{2}+4 + c |
Problem :
Evaluate
4x cos(x
^{2}+2)dx.
u | = x ^{2} +2 | ||
= 2x | |||
dx | = | ||
4x cos(u) | = 2cos(u)du = 2 sin(u) | ||
= 2 sin(x ^{2} + 2) + c |
Problem :
Evaluate
(x
)dx.
u | = x ^{2} -5 | ||
= 2x | |||
dx | = | ||
x | = du = = u ^{ } | ||
= x ^{2}-5 + c |
Problem :
Evaluate
(3x-4)^{2}
dx.
u = 3x - 4 | |||
= 3 | |||
dx = du | |||
u ^{2} du |
u ^{2} du = u ^{3} |
= 3x-4 _{0} ^{1} | |||
= 3(1)-4-(-4)^{3} | |||
= (-1)-(-64) | |||
= = 7 |
Problem :
Use the trapezoid rule with five subdivisions to approximate the area under
f (x) = x
^{2} + 1
on
[1, 6]
.
Area | 2+2(5)+2(10+2(17)+2(26)+37(1) | ||
= 77.5 |
16x2+1 dx=75 |