MATH 1210-1: Quiz 9A
Solution
Please show your work and MARK your answer.
1. (5 points) Find the following derivative:
d
dx
Z
x3
sin(t2 ) dt
x
Solution: By the first fundamental theorem of calculus, we have the following variant formula:
d
dx
Z
u(x)
f (t) dt = f (u(x))u0 (x) − f (l(x))l0 (x)
l(x)
And in this problem, f (x) = sin(t2 ), l(x) = x and u(x) = x3 . So
d
dx
Z
x3
sin(t2 ) dt = sin((x3 )2 ) · 3x2 − sin(x2 ) · 1
x
= 3x2 sin(x6 ) − sin(x2 )
2. (5 points) Evaluate the following definite integral:
Z
7
√
2
1
dx
x+2
Solution: Let u = x + 2, then x = u − 2 and du = dx. And notice that
u = 4 when x = 2 and u = 9 when x = 7.
Z 7
Z 9
√ ¯9
1
1
√
√ du = 2 u¯4 = 6 − 4 = 2
dx =
u
x+2
2
4
MATH 1210-1: Quiz 9B
Solution
Please show your work and MARK your answer.
1. (5 points) Find the following derivative:
d
dx
Z
x2
sin(t2 ) dt
x
Solution: By the first fundamental theorem of calculus, we have the following variant formula:
d
dx
Z
u(x)
f (t) dt = f (u(x))u0 (x) − f (l(x))l0 (x)
l(x)
And in this problem, f (x) = sin(t2 ), l(x) = x and u(x) = x2 . So
d
dx
Z
x2
sin(t2 ) dt = sin((x2 )2 ) · 2x − sin(x2 ) · 1
x
= 2x sin(x4 ) − sin(x2 )
2. (5 points) Evaluate the following definite integral:
Z
6
√
2
1
dx
x−2
Solution: Let u = x − 2, then x = u + 2 and du = dx. And notice that
u = 0 when x = 2 and u = 4 when x = 6.
Z 6
Z 4
√ ¯4
1
1
√
√ du = 2 u¯0 = 4 − 0 = 4
dx =
u
x−2
2
0