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AP Calculus BC
Unit 6: Integration and Accumulation of Change
6.11 Integrating Using Integration by Parts
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Integration by Parts is a technique used to integrate products of
functions
Steps to apply Integration by Parts
1️⃣ Choose
u
u
u
and
d
v
dv
d
v
2️⃣ Find
d
u
du
d
u
and
v
v
v
3️⃣ Apply the Formula
4️⃣ Evaluate the new integral
The Integration by Parts formula transforms
complex
integrals into simpler ones.
What is the Integration by Parts formula?
∫
u
d
v
=
\int u \, dv =
∫
u
d
v
=
u
v
−
∫
v
d
u
uv - \int v \, du
uv
−
∫
v
d
u
In the Integration by Parts formula, du</latex> is the derivative of
u
u
u
with respect to x
Why is choosing
u
u
u
and
d
v
dv
d
v
correctly important in Integration by Parts?
To simplify the new integral
The new integral in Integration by Parts is always more complex than the original.
False
The final step in Integration by Parts is to evaluate the new
integral
What is the formula for Integration by Parts?
∫
u
d
v
=
\int u \, dv =
∫
u
d
v
=
u
v
−
∫
v
d
u
uv - \int v \, du
uv
−
∫
v
d
u
Integration by Parts is based on the
product rule
of differentiation.
In Integration by Parts,
d
v
dv
d
v
is typically chosen so that it is easily integrable
What is the Integration by Parts formula?
∫
u
d
v
=
\int u \, dv =
∫
u
d
v
=
u
v
−
∫
v
d
u
uv - \int v \, du
uv
−
∫
v
d
u
What is the Integration by Parts formula?
∫
u
d
v
=
\int u \, dv =
∫
u
d
v
=
u
v
−
∫
v
d
u
uv - \int v \, du
uv
−
∫
v
d
u
By choosing
u
u
u
and
d
v
dv
d
v
appropriately, the new integral
∫
v
d
u
\int v \, du
∫
v
d
u
is often easier to solve than the original
Steps to solve an integral using Integration by Parts
1️⃣ Choose
u
u
u
and
d
v
dv
d
v
2️⃣ Find
d
u
du
d
u
and
v
v
v
3️⃣ Apply the formula
4️⃣ Evaluate the new integral
dv in the
Integration by Parts
formula should be easily integrable.
To find
d
u
du
d
u
, you need to differentiate
u
u
u
, and to find
v
v
v
, you need to integrate <dv</latex><dv</latex>
What is the result of choosing
u
=
u =
u
=
x
x
x
and
d
v
=
dv =
d
v
=
cos
(
x
)
d
x
\cos(x) \, dx
cos
(
x
)
d
x
in
∫
x
cos
(
x
)
d
x
\int x \cos(x) \, dx
∫
x
cos
(
x
)
d
x
?
−
x
cos
(
x
)
+
- x\cos(x) +
−
x
cos
(
x
)
+
sin
(
x
)
+
\sin(x) +
sin
(
x
)
+
C
C
C
Choosing
u
u
u
as the function that simplifies upon differentiation is a good strategy.
Match the function with its derivative or integral:
u = x ↔️ du = dx
dv = e^{x} dx ↔️ v = e^{x}
What is the value of
∫
e
x
d
x
\int e^{x} dx
∫
e
x
d
x
?
e^{x} + C</latex>
Integration by Parts is used to simplify
integrals
of products of functions.
Steps to solve an integral using Integration by Parts
1️⃣ Choose
u
u
u
and
d
v
dv
d
v
2️⃣ Find
d
u
du
d
u
and
v
v
v
3️⃣ Apply the formula
4️⃣ Evaluate the new integral
What is the Integration by Parts formula?
∫
u
d
v
=
\int u \, dv =
∫
u
d
v
=
u
v
−
∫
v
d
u
uv - \int v \, du
uv
−
∫
v
d
u
The new integral
∫
v
d
u
\int v \, du
∫
v
d
u
in Integration by Parts is often easier to solve than the original.
What is the Integration by Parts formula?
∫
u
d
v
=
\int u \, dv =
∫
u
d
v
=
u
v
−
∫
v
d
u
uv - \int v \, du
uv
−
∫
v
d
u
In the Integration by Parts formula,
u
u
u
and
v
v
v
are functions of x
The term
d
u
du
d
u
in the Integration by Parts formula represents the derivative of
u
u
u
with respect to
x
x
x
.
What does
d
v
dv
d
v
represent in the Integration by Parts formula?
The differential of
v
v
v
Steps for applying Integration by Parts
1️⃣ Choose
u
u
u
and
d
v
dv
d
v
2️⃣ Find
d
u
du
d
u
and
v
v
v
3️⃣ Apply the formula
4️⃣ Evaluate the new integral
For the integral
∫
x
e
x
d
x
\int x e^{x} dx
∫
x
e
x
d
x
, we choose
u
=
u =
u
=
x
x
x
and dv = e^{x} dx</latex>, which simplifies to dx
What is the derivative of
u
=
u =
u
=
x
x
x
in the example
∫
x
e
x
d
x
\int x e^{x} dx
∫
x
e
x
d
x
?
d
u
=
du =
d
u
=
d
x
dx
d
x
What is the integral of dv = e^{x} dx</latex> in the example
∫
x
e
x
d
x
\int x e^{x} dx
∫
x
e
x
d
x
?
v
=
v =
v
=
e
x
e^{x}
e
x
The final result of
∫
x
e
x
d
x
\int x e^{x} dx
∫
x
e
x
d
x
is
x
e
x
−
e
x
+
x e^{x} - e^{x} +
x
e
x
−
e
x
+
C
C
C
.
What does the acronym LIATE stand for in the context of Integration by Parts?
Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential
When using LIATE, the function that appears earlier in the list is chosen as
u
The acronym LIATE helps in choosing
d
v
dv
d
v
as the function that is easier to differentiate.
False
In the integral
∫
x
cos
(
x
)
d
x
\int x \cos(x) \, dx
∫
x
cos
(
x
)
d
x
, which function should be chosen as
u
u
u
?
u
=
u =
u
=
x
x
x
How is
d
u
du
d
u
found in the Integration by Parts process?
By differentiating
u
u
u
When finding
v
v
v
, no constant of integration is needed because it cancels out in the final formula
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