SME
Calculus II
Integration by parts, partial fractions, trig substitution
Use this for “Evaluate the integral of xe⁻ˣ from 0 to 2.”
Chart
Shaded: ∫ from 0 to 1 of f(x) dx = 2.097
All of it is above the axis, so the integral is the shaded area
This is an example. Type your own values to start a new problem.
∫
xe
x
dx
=
I
F(a)
Antiderivative at the lower limit
worked out (not typed)
F(b)
Antiderivative at the upper limit
worked out (not typed)
Formulas
F(
a
)
=
e
k
×
a
×
k
×
a
− 1
k
2
−0.25
=
e
2 × 0
×
2 × 0 − 1
2
2
F(
b
)
=
e
k
×
b
×
k
×
b
− 1
k
2
1.8473
=
e
2 × 1
×
2 × 1 − 1
2
2
I
=
F(
b
)
−
F(
a
)
2.0973 = 1.8473 − (−0.25)
Assumptions
•
Integration by parts: ∫u dv = uv − ∫v du. It trades the integral of a product for one that is easier.
•
Pick u = x, which gets simpler when differentiated (du = dx), and dv = ekx dx, so v = (1 ÷ k)ekx. LIATE is a rule of thumb for the choice.
•
Then ∫x·ekx dx = (x ÷ k)ekx − (1 ÷ k²)ekx, or F(x) = ekx(kx − 1) ÷ k², and the integral from a to b is F(b) − F(a).
•
The rate k cannot be 0, and area below the x-axis counts as negative.
Step-by-step
WE KNOW
k = 2, a = 0, b = 1
FIND
antiderivative at the lower limit (F(a)), antiderivative at the upper limit (F(b)), value of the integral (I)
Step 1 · Find antiderivative at the lower limit (F(a))
Use F(a) = ek × a × (k × a − 1) ÷ k²
By
parts
with
u
=
x
and
dv
=
e
kx
dx:
du
=
dx
and
v
=
(
1
k
)
e
kx
.
So
uv
−
∫v
du
gives
F(x)
=
e
kx
kx − 1
k
2
.
Put
in
x
=
a
.
F(
a
)
=
e
2 × 0
×
2 × 0 − 1
2
2
∫
e
2x
dx
=
(
1
2
)
e
2x
∫
x·
e
2x
dx
=
e
2x
2x − 1
4
+
C
F(
a
)
=
1
×
−1
4
F(
a
)
=
−0.25
Step 2 · Find antiderivative at the upper limit (F(b))
Use F(b) = ek × b × (k × b − 1) ÷ k²
The
same
antiderivative,
at
x
=
b
.
F(
b
)
=
e
2 × 1
×
2 × 1 − 1
2
2
F(
b
)
=
e
2
2
2
F(
b
)
=
7.389
4
F(
b
)
=
1.8473
Step 3 · Find value of the integral (I)
Use I = F(b) − F(a)
Subtract
F(
a
)
from
F(
b
),
by
the
Fundamental
Theorem.
I
=
1.8473
−
(−0.25)
1
∫
0
(
x
·
e
2
x
) dx
=
[
e
2x
2x − 1
4
]
from
0
to
1
=
2.0973
I
=
2.0973
Check
Put the numbers back in. Both sides should match.
−0.25
=
e
2 × 0
×
2 × 0 − 1
2
2
✓
1.8473
=
e
2 × 1
×
2 × 1 − 1
2
2
✓
2.0973 = 1.8473 − (−0.25) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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