SME
Calculus I · Problem type
An initial-value problem with f′ = A·eᵏˣ
Use this for “Find f if f′(x) = 6e²ˣ and f(0) = 5; then find f(1).”
Chart
f(1) ≈ 2.729
This is an example. Type your own values to start a new problem.
f′(x)
=
A
e
x
f(0)
=
y₀
f(
x₁
)
=
f(x₁)
C
Constant of integration
worked out (not typed)
Formulas
y₀
=
A
k
+
C
1
=
4
−2
+
3
f
(
x₁
)
=
(
A
k
)
×
e
k
×
x₁
+
C
2.7293
=
(
4
−2
)
×
e
−2 × 1
+
3
Assumptions
•
Every antiderivative of A·eᵏˣ is (A ÷ k)eᵏˣ + C: differentiating (1 ÷ k)eᵏˣ brings down the factor k (chain rule), which cancels the 1 ÷ k.
•
The value f(0) = y₀ fixes C: at x = 0, eᵏˣ = 1, so y₀ = A ÷ k + C.
•
The rate k is not 0. With k = 0, f′ is the constant A, and f(x) = Ax + y₀ instead.
Step-by-step
WE KNOW
A = 4, k = −2, y₀ = 1, x₁ = 1
FIND
constant of integration (C), value of f at x₁ (f(x₁))
Step 1 · Find constant of integration (C)
Use y₀ = A ÷ k + C
Antidifferentiate:
f
(x)
=
(
A
k
)eᵏˣ
+
C
.
At
x
=
0,
e
0
=
1,
so
f
(0)
=
A
k
+
C
;
take
A
k
away
from
y₀
.
C
=
y₀
−
A
k
C
=
1
−
4
−2
f
(x)
=
−2
e
−2x
+
C
f
(0)
=
−2
+
C
=
1
C
=
3
Step 2 · Find value of f at x₁ (f(x₁))
Use f(x₁) = (A ÷ k) × ek × x₁ + C
Now
that
C
is
known,
put
x₁
into
f
(x).
f
(
x₁
)
=
(
4
−2
)
×
e
−2 × 1
+
3
f
(
x₁
)
=
−2
×
e
-2
+
3
f
(
x₁
)
=
−2
×
0.1353353
+
3
f
(
x₁
)
=
−0.2706706
+
3
f
(
x₁
)
=
2.7293
Check
Put the numbers back in. Both sides should match.
1
=
4
−2
+
3
✓
2.7293
=
(
4
−2
)
×
e
−2 × 1
+
3
✓
Topic check
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