Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus I
April 5, 2016
Name:
Quiz 17
Derivatives of integrals Simplify the following expression
Z x3
d
1
dp
dx 2 p2
Recall: For any n with n 6= −1 we have
b
Z b
Z
1
1
1
pn dp =
pn+1 , and
pn+1 =
(bn+1 − an+1 ),
(1)
pn dp =
n+1
n
+
1
n
+
1
a
a
(2)
d n
x = n · xn−1 ,
dx
d
dx
Z
2
x3
1
dp
p2
d
c = 0 where c is a number.
dx
x3
d
dx
Z
(1)
d
dx
simplify
d
dx
=
=
=
(1)
=
=
=
(2)
p−2 dp
2
x3
1
−2+1 p
−2 + 1
2
x3
− p−1 2
d
− (x3 )−1 − (−2)−1
dx
d
1
−3
−x +
dx
2
d
d 1
−3
−x
+
dx
dx 2
=
−(−3)x−3−1
=
3
x4