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Calculus AB · Unit 6 · 6.4
AP Calculus · Integration and Accumulation of Change
The Fundamental Theorem and Accumulation Functions
Differentiating an integral hands the integrand back — as long as you respect the upper limit.
key idea$g(x)=\int_a^x f(t)\,dt\ \Rightarrow\ g'(x)=f(x)$
The theorem
The Fundamental Theorem of Calculus (accumulation form)
Hypotheses — all must hold
- $f$ is continuous on an interval containing $a$.
- $g$ is defined by $g(x)=\displaystyle\int_a^x f(t)\,dt$ — the accumulated area from the fixed lower limit $a$ to the variable upper limit $x$.
What you may then conclude
$g$ is differentiable and $g'(x)=f(x)$. Accumulating and differentiating are inverse operations — differentiating an integral hands the integrand straight back.
The upper limit must be $x$ itself
If the upper limit is a function of $x$, the chain rule intervenes: $\dfrac{d}{dx}\displaystyle\int_a^{u(x)}f(t)\,dt=f\!\left(u(x)\right)\cdot u'(x)$. Reporting just $f\!\left(u(x)\right)$ is the single most common error on this topic.
$g(x)=\int_0^x f(t)\,dt$ accumulates this area
Checking the hypotheses
| Accumulation function | Its derivative | Why |
|---|
| $\displaystyle\int_a^x f(t)\,dt$ | $f(x)$ | the theorem, directly |
| $\displaystyle\int_a^{x^{2}} f(t)\,dt$ | $f\!\left(x^{2}\right)\cdot 2x$ | chain rule on the upper limit |
| $\displaystyle\int_x^b f(t)\,dt$ | $-f(x)$ | reverse the limits first, picking up a minus |
| $\displaystyle\int_{u(x)}^{v(x)} f(t)\,dt$ | $f(v)v'-f(u)u'$ | split at a constant, then chain rule both ends |
| $\displaystyle\int_a^b f(t)\,dt$ | $0$ | a constant — no variable limit |
Row 5 catches people out: with both limits constant the integral is a number, so its derivative is zero, whatever $f$ is.
reading g from f
- $g$ accumulates signed area — it decreases where $f$ is negative.
- $g(a)=0$ always, because the interval has zero width there.
- $g'=f$, so extrema of $g$ sit at zeros of $f$, and concavity of $g$ follows the sign of $f'$.
- The variable of integration $t$ is internal; the answer is a function of $x$.
Worked examples
EX 1
Differentiate an accumulation function
directFind $g'(x)$ for $g(x)=\displaystyle\int_{1}^{x}\sqrt{t^{3}+1}\,dt$.
1
Check the form
Lower limit constant, upper limit exactly $x$, integrand continuous. The theorem applies directly.
2
Apply it
$g'(x)=\sqrt{x^{3}+1}$.
3
Note what was avoided
No antiderivative of $\sqrt{t^{3}+1}$ exists in elementary terms — and none was needed.
4
Read the notation
The $t$ vanished because it is internal. Replace it with $x$ and you have the answer.
EX 2
A function as the upper limit
chain ruleFind $\dfrac{d}{dx}\displaystyle\int_{0}^{x^{2}}\cos(t)\,dt$.
1
Spot the composition
The upper limit is $u(x)=x^{2}$, not $x$.
2
Apply the extended form
$f\!\left(u(x)\right)\cdot u'(x)$.
3
Substitute
$\cos\!\left(x^{2}\right)\cdot 2x$.
4
Present it
$2x\cos\!\left(x^{2}\right)$.
5
Verify independently
$\int_0^{x^{2}}\cos t\,dt=\sin\!\left(x^{2}\right)$, whose derivative is $2x\cos\!\left(x^{2}\right)$. ✓
EX 3
Analyse $g$ from a graph of $f$
reading$g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$ where $f\gt 0$ on $(0,3)$, $f\lt 0$ on $(3,7)$, and $f(3)=0$. Where does $g$ have its maximum?
1
Use the theorem
$g'=f$, so the critical points of $g$ are the zeros of $f$.
2
Locate them
$f(3)=0$, so $x=3$ is the only critical point on $(0,7)$.
3
Apply the First Derivative Test
$g'=f$ changes from positive to negative at $x=3$.
4
Conclude
$g$ has a local — and here absolute — maximum at $x=3$.
5
Interpret
Area was accumulating while $f$ was above the axis; past $x=3$ the negative region starts subtracting.
EX 4
Both limits variable
the general formFind $\dfrac{d}{dx}\displaystyle\int_{x}^{x^{3}}e^{t^{2}}\,dt$.
1
Split at any constant
$\displaystyle\int_{x}^{x^{3}}=\int_{x}^{0}+\int_{0}^{x^{3}}=-\int_{0}^{x}+\int_{0}^{x^{3}}$.
2
Differentiate the second piece
Upper limit $x^{3}$: $e^{\left(x^{3}\right)^{2}}\cdot3x^{2}=3x^{2}e^{x^{6}}$.
3
Differentiate the first piece
$-\displaystyle\int_{0}^{x}$ differentiates to $-e^{x^{2}}$.
4
Combine
$3x^{2}e^{x^{6}}-e^{x^{2}}$.
5
State the pattern
$\dfrac{d}{dx}\displaystyle\int_{u}^{v}f=f(v)v'-f(u)u'$ — top minus bottom, each with its own chain factor.
When it fails
The chain factor on a non-$x$ upper limit
$\dfrac{d}{dx}\int_0^{x^{2}}f=f\!\left(x^{2}\right)\cdot\mathbf{2x}$. Omitting the $2x$ is the standard error.
Constant limits give a constant
$\int_a^b f(t)\,dt$ is a number, so its derivative is $0$ — regardless of how complicated $f$ is.
A variable in the lower limit contributes a minus
Reverse the limits first. $\dfrac{d}{dx}\int_x^b f=-f(x)$.
Continuity is required
If $f$ has a discontinuity between $a$ and $x$, the accumulation function need not be differentiable there.