SME
Calculus II · Problem type
An exponential decay to infinity
Use this for “Evaluate the integral of e⁻³ˣ from 0 to ∞, or show it diverges.”
Chart
Shaded: ∫ from 0 to 10 of f(x) dx = 1.987
This is an example. Type your own values to start a new problem.
∫
0
e
−
x
dx
=
Ib
∫
∞
0
e
−
x
dx
=
I
Formulas
Ib
=
1 −
e
−
k
×
b
k
1.9865
=
1 −
e
−0.5 × 10
0.5
I
=
1
k
2
=
1
0.5
Assumptions
•
An improper integral is a limit of ordinary ones: ∫ from 0 to ∞ of e−kx dx is the limit of ∫ from 0 to b as b → ∞.
•
An antiderivative of e−kx is −(1 ÷ k)e−kx, so the integral from 0 to b is (1 − e−kb) ÷ k.
•
The rate k is more than 0, so e−kb shrinks to 0 as b grows and the integral converges to 1 ÷ k.
Step-by-step
WE KNOW
k = 0.5, b = 10
FIND
integral from 0 to b (Ib), limit as b → ∞ (I)
Step 1 · Find integral from 0 to b (Ib)
Use Ib = (1 − e−k × b) ÷ k
Integrate
e
−kx
to
−(
1
k
)
e
−kx
.
Take
its
value
at
0,
which
is
−
1
k
,
from
its
value
at
b
.
Ib
=
1 −
e
−0.5 × 10
0.5
∫
e
−0.5x
dx
=
−(
1
0.5
)
e
−0.5x
+
C
10
∫
0
e
−0.5
x
dx
=
[−(
1
0.5
)
e
−0.5x
]
from
0
to
10
=
1.9865
Ib = 1.9865
Step 2 · Find limit as b → ∞ (I)
Use I = 1 ÷ k
As
b
→
∞,
e
−
kb
→
0
because
k
>
0,
so
the
integral
to
b
approaches
1 − 0
k
=
1
k
.
It
converges.
I
=
1
0.5
I
=
2
Check
Put the numbers back in. Both sides should match.
1.9865
=
1 −
e
−0.5 × 10
0.5
✓
2
=
1
0.5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.