SME
Calculus I
u-substitution
Use this for “Evaluate the integral of x(x² − 3)² from 1 to 2.”
Chart
Shaded: ∫ from 0 to 1 of f(x) dx = 1.875
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.
∫
x
(x²
+
)
dx
=
15
8
I
u(a)
New lower limit, a² + c
worked out (not typed)
u(b)
New upper limit, b² + c
worked out (not typed)
Formulas
u(
a
)
=
a
2
+
c
1
=
0
2
+
1
u(
b
)
=
b
2
+
c
2
=
1
2
+
1
I
=
(u
(
b
)
n
+ 1
−
u
(
a
)
n
+ 1
)
÷
(2
×
(
n
+
1))
15
8
=
2
3 + 1
−
1
3 + 1
2 × (3 + 1)
Assumptions
•
Pick u as the inside part, u = x² + c. Its derivative 2x is already outside, up to a factor of 2, so x dx = du ÷ 2.
•
Change the limits with u: x = a becomes u(a) = a² + c and x = b becomes u(b) = b² + c. Then there is no need to go back to x.
•
In u the integral is ½ × ∫ uⁿ du, and by the power rule an antiderivative is uⁿ⁺¹ ÷ (2(n + 1)).
•
Area below the x-axis counts as negative, as for any definite integral.
Step-by-step
WE KNOW
c = 1, n = 3, a = 0, b = 1
FIND
new lower limit, a² + c (u(a)), new upper limit, b² + c (u(b)), value of the integral (I)
Step 1 · Find new lower limit, a² + c (u(a))
Use u(a) = a² + c
Put
the
lower
limit
x
=
a
into
the
inside,
u
=
x
2
+
c
:
the
integral
in
u
starts
there.
u(
a
)
=
0
2
+
1
u(
a
)
=
1
Step 2 · Find new upper limit, b² + c (u(b))
Use u(b) = b² + c
Put
the
upper
limit
x
=
b
into
the
inside:
the
integral
in
u
ends
there.
u(
b
)
=
1
2
+
1
u(
b
)
=
1
+
1
u(
b
)
=
2
Step 3 · Find value of the integral (I)
Use I = (u(b)ⁿ⁺¹ − u(a)ⁿ⁺¹) ÷ (2 × (n + 1))
Since
du
=
2x
dx,
x
dx
is
du
2
:
the
integral
becomes
half
the
integral
of
uⁿ
from
u(
a
)
to
u(
b
).
Raise
the
power
by
1,
then
divide
by
the
new
power
and
by
2.
I
=
2
4
−
1
4
8
u
=
x
2
+
1,
du
=
2x
dx,
so
x
dx
=
du
2
1
∫
0
(
x
(
x
2
+ 1)
3
) dx
=
2
∫
1
(
u
3
2
) du
2
∫
1
(
u
3
2
) du
=
[
u
4
8
]
from
1
to
2
=
15
8
I
=
15
8
Check
Put the numbers back in. Both sides should match.
1
=
0
2
+
1
✓
2
=
1
2
+
1
✓
15
8
=
2
3 + 1
−
1
3 + 1
2 × (3 + 1)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.