SME
Calculus I · Problem type
Area between a line and a parabola
Use this for “Find the area of the region between y = 2x + 3 and y = x².”
Chart
f(x) = g(x)
The curves cross at x = −1 and x = 2.
Shaded between f and g from −1 to 2: area ∫|f − g| dx = 4.5
This is an example. Type your own values to start a new problem.
∫
(
m
x
+
c
−
x²)
dx
=
9
2
A
Formulas
x₁
=
m
−
√
m
2
+ 4 ×
c
2
−1
=
1 −
√
1
2
+ 4 × 2
2
x₂
=
m
+
√
m
2
+ 4 ×
c
2
2
=
1 +
√
1
2
+ 4 × 2
2
A
=
(
x₂
−
x₁
)
3
÷
6
9
2
=
(2 − (−1))³
6
Assumptions
•
The line y = mx + c and the parabola y = x² meet where x² = mx + c, at the crossings x₁ and x₂. Between them the line is on top.
•
A thin strip at x is as tall as the top curve less the bottom one, (mx + c) − x², so A = ∫ from x₁ to x₂ of (mx + c − x²) dx.
•
Top less bottom keeps every strip positive, so the area is never negative, even where the region lies below the x-axis.
•
A shortcut: mx + c − x² = (x − x₁)(x₂ − x), and its integral from x₁ to x₂ is (x₂ − x₁)³ ÷ 6.
Step-by-step
WE KNOW
m = 1, c = 2
FIND
left crossing of the curves (x₁), right crossing of the curves (x₂), area between the line and the parabola (A)
Step 1 · Find left crossing of the curves (x₁)
Use x₁ = (m − √(m² + 4 × c)) ÷ 2
Set
the
curves
equal:
x
2
=
mx
+
c
,
so
x
2
−
mx
−
c
=
0.
The
quadratic
formula’s
smaller
root
is
the
left
crossing.
x₁
=
1 −
√
1
2
+ 4 × 2
2
x₁
=
1 −
√
1 + 4 × 2
2
x₁
=
1 −
√
1 + 8
2
x₁
=
1 −
√
9
2
x₁
=
1 − 3
2
x₁
=
−
2
2
x₁
=
−1
Step 2 · Find right crossing of the curves (x₂)
Use x₂ = (m + √(m² + 4 × c)) ÷ 2
The
larger
root
of
x
2
−
mx
−
c
=
0
is
the
right
crossing.
x₂
=
1 +
√
1
2
+ 4 × 2
2
x₂
=
1 +
√
1 + 4 × 2
2
x₂
=
1 +
√
1 + 8
2
x₂
=
1 +
√
9
2
x₂
=
1 + 3
2
x₂
=
4
2
x₂
=
2
Step 3 · Find area between the line and the parabola (A)
Use A = (x₂ − x₁)³ ÷ 6
Integrate
the
line
less
the
parabola,
x
+
2
−
x
2
,
from
x₁
to
x₂
:
an
antiderivative
is
−
x
3
3
+
x
2
2
+
2x.
It
comes
to
(
x₂
−
x₁
)
3
÷
6.
A
=
(2 − (−1))³
6
2
∫
−1
(
x
+ 2 −
x
2
) dx
=
[−
x
3
3
+
x
2
2
+
2x]
from
−1
to
2
=
9
2
A
=
9
2
Check
Put the numbers back in. Both sides should match.
−1
=
1 −
√
1
2
+ 4 × 2
2
✓
2
=
1 +
√
1
2
+ 4 × 2
2
✓
9
2
=
(2 − (−1))³
6
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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