SME
Calculus I
Antiderivatives and initial-value problems
Use this for “Find f if f′(x) = 6x² − 4x + 1 and f(1) = 5. Then find f(2).”
Chart
f(2) = 14
This is an example. Type your own values to start a new problem.
f′(x)
=
p
x²
+
(
q
)
x
+
r
f(
x₀
)
=
y₀
f(
x₁
)
=
f(x₁)
C
Constant of integration
worked out (not typed)
Formulas
y₀
=
p
×
x₀
3
3
+
q
×
x₀
2
2
+
r
×
x₀
+
C
5
=
6
×
1
3
3
+
(−4)
×
1
2
2
+
1
×
1
+
4
f
(
x₁
)
=
p
×
x₁
3
3
+
q
×
x₁
2
2
+
r
×
x₁
+
C
14
=
6
×
2
3
3
+
(−4)
×
2
2
2
+
1
×
2
+
4
Assumptions
•
Every antiderivative of px² + qx + r is px³ ÷ 3 + qx² ÷ 2 + rx + C: raise each power by 1, divide by the new power, and add a constant C.
•
Two antiderivatives of the same f′ differ only by a constant, so one point (x₀, y₀) on the curve fixes C.
•
Check the answer by differentiating it back: the slope of f must be f′, and f(x₀) must be y₀.
Step-by-step
WE KNOW
p = 6, q = −4, r = 1, x₀ = 1, y₀ = 5, x₁ = 2
FIND
constant of integration (C), value of f at x₁ (f(x₁))
Step 1 · Find constant of integration (C)
Use y₀ = p × x₀³ ÷ 3 + q × x₀² ÷ 2 + r × x₀ + C
Antidifferentiate
term
by
term:
f
(x)
=
2x³
−
2x²
+
x
+
C
.
Put
in
x
=
x₀
,
set
it
equal
to
y₀
and
take
the
rest
away.
C
=
5
−
(6
×
1
3
3
−
4
×
1
2
2
+
1)
f
(x)
=
2x³
−
2x²
+
x
+
C
f
(1)
=
1
+
C
=
5
C
=
4
Step 2 · Find value of f at x₁ (f(x₁))
Use f(x₁) = p × x₁³ ÷ 3 + q × x₁² ÷ 2 + r × x₁ + C
Now
that
C
is
known,
put
x₁
into
f
(x).
f
(
x₁
)
=
6
×
2
3
3
−
4
×
2
2
2
+
2
+
4
f
(
x₁
)
=
14
Check
Put the numbers back in. Both sides should match.
5
=
6
×
1
3
3
+
(−4)
×
1
2
2
+
1
×
1
+
4
✓
14
=
6
×
2
3
3
+
(−4)
×
2
2
2
+
1
×
2
+
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.