Problem. Let f(x) and g(x) be two functions such that:
R2
R2
R0
[f (x) + g(x)]dx = 3, −1 [f (x) − 2g(x)]dx = 1, −1 f (x)dx = −1
−1
R2
Find 0 f (x)dx
Proof.
Z
2
f (x)dx =
0
=
Z
2
−1
Z 2
f (x)dx −
Z
0
f (x)dx
(1)
−1
f (x)dx − (−1)
(2)
−1
and we have:
Z
2
Z 2
Z 2
1
2
[f (x) + g(x)]dx +
[f (x) − 2g(x)]dx
3
−1
−1
1
= (2 · 3 + 1)
3
7
=
3
f (x)dx =
−1
(3)
(4)
(5)
so
that:
R 2 we can conclude
7
10
f
(x)dx
=
+
1
=
3
3
0
2
d
Problem. 1) find dx
ex
2)Evaluate
R 2 x2
xe dx
0
Proof. 1) we use the chain rule to get:
d
dx
2
ex
2
= 2xex
2
2
2) Note that by part (1), the antiderivative of xex is 12 ex . Now apply the
FUNDAMENTAL THEOREM OF CALCULUS to get:
i
R 2 x2
2 2
xe dx = 21 ex
= 21 e4 − e0 = 21 e4 − 1
0
0
Problem. Find the antiderivative of:
1. sin(2x)
cos(x)
2. ex+7 2−2x
2
3. xx3
Proof. 1. Use the ‘double angle formula’: sin(2x) = 2sin(x)cos(x) and then
the Rproblem becomes
VERY simple: R
R 2sin(x)cos(x)
sin(2x)
dx = 2sin(x)dx = −2cos(x) + C
cos(x) dx =
cos(x)
2. So this problem seems like it may contain integration by part or substitution, but if we SIMPLIFY
wellwe can avoid both:
x
x
ex+7 2−2x = e7 ex 14 = e7 4e
1
Now there is only ONE function and it is NOT composite thus:
Z
Z
e x
x+7 −2x
dx
e
2
dx = e7
4
Z x
e
= e7
dx
4
1 e x
+C
= e7
4
ln 4e
e x
e7
+C
=
1 − ln(4) 4
(6)
(7)
(8)
(9)
3. If there is one thing you should be noticing by now, it is this: TRY TO
REDUCE THE FUNCTION BEFORE INTEGRATING OR DERIVATING!
This last one is very simple one line calculation if we just notice that:
x2
−1
(REDUCTION OF FRACTIONS HAS COME UP A LOT SO
x3 = x
PAY ATTENTION TO IT!)
And
R x2 so R −1
x dx = ln(x) + C
x3 dx =
2