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Calculus I · Problem type

The area function and the Fundamental Theorem, part 1

Use this for “F(x) is the integral of 3t − 2 from 1 to x. Find F(4) and F′(4).”

Chart

f(t) = 2t + 1−1012345−6−4−202468101214ty−10010203040F(x) = ∫ from 0 to x of f(t) dtF(3) = 12slope f(3) = 7(3, 7)
f(3) = 7
F(x) = ∫ from 0 to x of f(t) dt: F(3) = 12, and its slope there is f(3) = 7
This is an example. Type your own values to start a new problem.
F(
x
)
=
∫
(
m
t
+
c
)
dt
=
F(x)
F′(
x
)
=
F′(x)

Formulas

F(
x
)
=
m
×
x
2
−
a
2
2
+
c
×
(
x
−
a
)
12
=
2
×
3
2
−
0
2
2
+
1
×
(3
−
0)
F′(
x
)
=
m
×
x
+
c
7 = 2 × 3 + 1

Assumptions

•
F(x) = ∫ from a to x of f(t) dt collects the signed area under f(t) = mt + c from a up to x. The letter t runs along the axis; x is where it stops.
•
An antiderivative of mt + c is G(t) = mt² ÷ 2 + ct, so F(x) = G(x) − G(a), by the Fundamental Theorem (part 2).
•
The Fundamental Theorem (part 1): F′(x) = f(x). Moving x on adds a thin strip as tall as f(x), so the area grows at that rate.
•
Area below the t-axis counts as negative, and F(a) = 0. When x is left of a, F(x) is minus the area from x to a.

Step-by-step

WE KNOW
m = 2, c = 1, a = 0, x = 3
FIND
area function at x (F(x)), slope of the area function at x (F′(x))
Step 1 · Find area function at x (F(x))
Use F(x) = m × (x² − a²) ÷ 2 + c × (x − a)
An
antiderivative
of
f
(t)
=
2t
+
1
is
G(t)
=
t
2
+
t.
Take
G
at
x
less
G
at
a
:
m
(
x
2
−
a
2
)
2
+
c
(
x
−
a
).
F(
x
)
=
2
×
3
2
−
0
2
2
+
1
×
(3
−
0)
3
∫
0
(2
t
+ 1) dt
=
[
t
2
+
t]
from
0
to
3
=
12
F(
x
)
=
12
Step 2 · Find slope of the area function at x (F′(x))
Use F′(x) = m × x + c
By
the
Fundamental
Theorem
(part
1),
the
slope
of
F
at
x
is
the
height
of
f
at
x
:
put
t
=
x
into
mt
+
c
.
No
integral
is
needed.
F′(
x
)
=
2
×
3
+
1
F′(
x
)
=
6
+
1
F′(
x
)
=
7
Check
Put the numbers back in. Both sides should match.
12
=
2
×
3
2
−
0
2
2
+
1
×
(3
−
0)
✓
7 = 2 × 3 + 1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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