The Central Idea
Earlier, we studied two apparently different operations.
Differentiation:
\[ F(x)\longrightarrow F'(x). \]Integration:
\[ f(x)\longrightarrow \int_a^b f(x)\,dx. \]The Fundamental Theorem of Calculus reveals that these are inverse processes in a precise sense.
\[ \boxed{\text{differentiation and integration undo one another}}. \]The Theorem Has Two Main Parts
The Fundamental Theorem of Calculus is usually presented in two related statements.
- An accumulation function built from a continuous integrand has derivative equal to the integrand.
- A definite integral can be evaluated using any antiderivative of the integrand.
These are often called the first and second parts of the theorem.
Accumulation Functions
Let \(f\) be continuous and define
\[ A(x)=\int_a^x f(t)\,dt. \]The lower limit \(a\) is fixed, while the upper limit \(x\) varies.
As \(x\) moves, \(A(x)\) records the net accumulation of \(f\) from \(a\) to \(x\).
This makes \(A\) an accumulation function.
Why We Often Use \(t\) Inside the Integral
In
\[ A(x)=\int_a^x f(t)\,dt, \]the symbol \(x\) is already being used as the upper limit.
The variable \(t\) is used as a local variable of integration.
We could use another symbol instead:
\[ \int_a^x f(u)\,du. \]The value is unchanged.
Fundamental Theorem of Calculus, Part I
If \(f\) is continuous on an interval containing \(a\) and \(x\), and
\[ A(x)=\int_a^x f(t)\,dt, \]then
\[ \boxed{A'(x)=f(x)}. \]In expanded notation,
\[ \boxed{ \frac{d}{dx} \left[ \int_a^x f(t)\,dt \right] = f(x) }. \]What Part I Means
The function
\[ A(x)=\int_a^x f(t)\,dt \]measures accumulated value.
Part I says that the rate at which this accumulated value changes is the original function:
\[ \boxed{\text{rate of accumulation}=f(x)}. \]In other words, accumulation creates an antiderivative.
Why Part I Is Reasonable
Suppose \(x\) increases by a small amount \(\Delta x\).
The accumulation function changes by approximately
\[ \Delta A\approx f(x)\Delta x. \]Dividing by \(\Delta x\),
\[ \frac{\Delta A}{\Delta x}\approx f(x). \]As \(\Delta x\to0\), the average rate approaches the instantaneous rate:
\[ \boxed{A'(x)=f(x)}. \]Example: Differentiating an Accumulation Function
Let
\[ A(x)=\int_2^x (t^3+4t)\,dt. \]By the Fundamental Theorem of Calculus,
\[ \boxed{A'(x)=x^3+4x}. \]No explicit integration is required.
Example: Trigonometric Integrand
Suppose
\[ G(x)=\int_0^x \cos t\,dt. \]Then
\[ \boxed{G'(x)=\cos x}. \]Example: Exponential Accumulation
If
\[ H(x)=\int_1^x e^{t^2}\,dt, \]then
\[ \boxed{H'(x)=e^{x^2}}. \]Notice that \(e^{t^2}\) does not have an elementary antiderivative.
Part I still lets us differentiate the accumulation function immediately.
When the Upper Limit Is a Function
Suppose
\[ A(x)=\int_a^{g(x)}f(t)\,dt. \]The upper limit now changes according to \(g(x)\).
Apply the Fundamental Theorem first and then the chain rule:
\[ \boxed{ A'(x)=f(g(x))g'(x) }. \]Example: Variable Upper Limit
Differentiate
\[ A(x)=\int_0^{x^2}(1+t^4)\,dt. \]Evaluate the integrand at the upper limit:
\[ 1+(x^2)^4=1+x^8. \]Multiply by the derivative of \(x^2\):
\[ 2x. \]Therefore,
\[ \boxed{A'(x)=2x(1+x^8)}. \]Example: Radical Upper Limit
Differentiate
\[ F(x)=\int_3^{\sqrt{x}}\sin(t^2)\,dt. \]By the chain-rule form,
\[ F'(x) = \sin\!\left((\sqrt{x})^2\right) \frac{d}{dx}(\sqrt{x}). \]Thus,
\[ \boxed{ F'(x)=\frac{\sin x}{2\sqrt{x}} }. \]When the Lower Limit Is Variable
Suppose
\[ A(x)=\int_{g(x)}^b f(t)\,dt. \]Reverse the limits:
\[ A(x) = -\int_b^{g(x)}f(t)\,dt. \]Differentiate:
\[ \boxed{ A'(x) = -f(g(x))g'(x) }. \]Example: Variable Lower Limit
Differentiate
\[ G(x)=\int_{x^3}^{5}\cos t\,dt. \]The variable limit is lower, so the result receives a negative sign:
\[ G'(x) = -\cos(x^3)(3x^2). \]Therefore,
\[ \boxed{ G'(x)=-3x^2\cos(x^3) }. \]When Both Limits Are Variable
Suppose
\[ A(x)=\int_{u(x)}^{v(x)}f(t)\,dt. \]Split at any convenient fixed point \(c\):
\[ A(x) = \int_c^{v(x)}f(t)\,dt - \int_c^{u(x)}f(t)\,dt. \]Differentiating gives
\[ \boxed{ A'(x) = f(v(x))v'(x) - f(u(x))u'(x) }. \]Example: Two Variable Limits
Differentiate
\[ K(x)=\int_x^{x^2}(1+t^2)\,dt. \]The upper-limit contribution is
\[ [1+(x^2)^2](2x). \]The lower-limit contribution is
\[ -(1+x^2)(1). \]Therefore,
\[ \boxed{ K'(x)=2x(1+x^4)-(1+x^2) }. \]Fundamental Theorem of Calculus, Part II
Suppose \(f\) is continuous on \([a,b]\), and \(F\) is any antiderivative of \(f\):
\[ F'(x)=f(x). \]Then
\[ \boxed{ \int_a^b f(x)\,dx = F(b)-F(a) }. \]This is the principal computational form of the theorem.
Evaluation-Bar Notation
The expression
\[ F(b)-F(a) \]is often abbreviated as
\[ \boxed{ \left.F(x)\right|_a^b }. \]We also commonly write
\[ \left[F(x)\right]_a^b. \]Both mean:
\[ \boxed{\text{upper value minus lower value}}. \]Example: Evaluating a Definite Integral
Evaluate
\[ \int_1^3 2x\,dx. \]An antiderivative of \(2x\) is
\[ F(x)=x^2. \]Apply Part II:
\[ \int_1^3 2x\,dx = [x^2]_1^3. \]Therefore,
\[ 3^2-1^2=9-1. \]Hence,
\[ \boxed{8}. \]Example: Polynomial Integral
Evaluate
\[ \int_0^2(3x^2+4x-1)\,dx. \]An antiderivative is
\[ x^3+2x^2-x. \]Therefore,
\[ \int_0^2(3x^2+4x-1)\,dx = [x^3+2x^2-x]_0^2. \]Evaluate:
\[ (8+8-2)-0. \]Thus,
\[ \boxed{14}. \]Example: Logarithmic Antiderivative
Evaluate
\[ \int_1^e \frac{1}{x}\,dx. \]An antiderivative is
\[ \ln|x|. \]Since \(x>0\) on \([1,e]\),
\[ \int_1^e\frac1x\,dx = [\ln x]_1^e. \]Therefore,
\[ \ln e-\ln1=1-0. \]Hence,
\[ \boxed{1}. \]Example: Trigonometric Integral
Evaluate
\[ \int_0^{\pi/2}\cos x\,dx. \]An antiderivative is \(\sin x\), so
\[ [\sin x]_0^{\pi/2} = \sin\frac{\pi}{2}-\sin0. \]Therefore,
\[ \boxed{1}. \]Example: Exponential Integral
Evaluate
\[ \int_0^1 e^x\,dx. \]Since \(e^x\) is its own antiderivative,
\[ \int_0^1e^x\,dx = [e^x]_0^1. \]Therefore,
\[ \boxed{e-1}. \]Why the Constant of Integration Disappears
Suppose the antiderivative is
\[ F(x)+C. \]Then
\[ [F(x)+C]_a^b = (F(b)+C)-(F(a)+C). \]The constants cancel:
\[ F(b)-F(a). \]This is why no \(+C\) is needed when evaluating a definite integral.
Any Antiderivative Works
If \(F\) and \(G\) are both antiderivatives of \(f\), then
\[ G(x)=F(x)+C. \]Therefore,
\[ G(b)-G(a) = F(b)-F(a). \]So the definite integral does not depend on which member of the antiderivative family we choose.
Why Part II Is So Powerful
The definite integral was defined using a limit of sums:
\[ \lim_{n\to\infty} \sum_{i=1}^{n}f(x_i^*)\Delta x. \]Evaluating such limits directly can be difficult.
Part II replaces the entire limiting computation with:
\[ \boxed{\text{find an antiderivative and evaluate it at two endpoints}}. \]This is one of the major computational simplifications in mathematics.
The Net Change Theorem
If
\[ F'(x)=f(x), \]then the Fundamental Theorem gives
\[ \boxed{ \int_a^b F'(x)\,dx = F(b)-F(a) }. \]This says:
\[ \boxed{\text{accumulated rate of change}=\text{net change}}. \]Example: Velocity and Position
Since
\[ s'(t)=v(t), \]the net change theorem gives
\[ \int_{t_1}^{t_2}v(t)\,dt = s(t_2)-s(t_1). \]Therefore,
\[ \boxed{\text{displacement}=\text{final position}-\text{initial position}}. \]Example: Displacement from Velocity
Suppose
\[ v(t)=3t^2-4t+2 \]for \(0\leq t\leq2\).
Displacement is
\[ \int_0^2(3t^2-4t+2)\,dt. \]An antiderivative is
\[ t^3-2t^2+2t. \]Therefore,
\[ [t^3-2t^2+2t]_0^2 = 8-8+4. \]Hence,
\[ \boxed{4}. \]Example: Accumulated Volume from Flow Rate
Suppose water enters a tank at rate
\[ r(t)=2t+5 \]liters per minute.
The volume entering from \(t=1\) to \(t=4\) is
\[ \int_1^4(2t+5)\,dt. \]An antiderivative is
\[ t^2+5t. \]Therefore,
\[ [t^2+5t]_1^4 = (16+20)-(1+5). \]Hence,
\[ \boxed{30\text{ liters}}. \]Example: Marginal Cost and Total Cost Change
Suppose marginal cost is
\[ C'(q)=0.02q+5. \]The change in cost from \(q=100\) to \(q=200\) is
\[ \int_{100}^{200}(0.02q+5)\,dq. \]An antiderivative is
\[ 0.01q^2+5q. \]Therefore,
\[ [0.01q^2+5q]_{100}^{200}. \]This gives
\[ (400+1000)-(100+500). \]Hence,
\[ \boxed{800}. \]Evaluating an Accumulation Function Explicitly
Suppose
\[ A(x)=\int_1^x 2t\,dt. \]By Part II,
\[ A(x) = [t^2]_1^x. \]Therefore,
\[ \boxed{A(x)=x^2-1}. \]Differentiating gives
\[ A'(x)=2x, \]exactly as Part I predicts.
Seeing Both Parts Together
Let
\[ A(x)=\int_0^x 3t^2\,dt. \]By Part I,
\[ A'(x)=3x^2. \]By Part II,
\[ A(x)=[t^3]_0^x=x^3. \]Differentiate:
\[ \frac{d}{dx}(x^3)=3x^2. \]The two parts agree perfectly.
Accumulating in the Opposite Direction
Consider
\[ B(x)=\int_x^a f(t)\,dt. \]Since
\[ B(x)=-\int_a^x f(t)\,dt, \]differentiation gives
\[ \boxed{B'(x)=-f(x)}. \]The negative sign reflects the reversed orientation.
How the Integrand Controls the Accumulation Function
Since
\[ A'(x)=f(x), \]the sign of \(f\) determines whether \(A\) increases or decreases.
- If \(f(x)>0\), then \(A\) is increasing.
- If \(f(x)<0\), then \(A\) is decreasing.
- If \(f(x)=0\), then \(A\) has zero instantaneous slope.
How \(f'\) Controls the Concavity of Accumulation
Because
\[ A'(x)=f(x), \]differentiating again gives
\[ A''(x)=f'(x). \]Therefore:
- if \(f\) is increasing, then \(A\) is concave up;
- if \(f\) is decreasing, then \(A\) is concave down.
This gives a powerful way to analyze accumulation functions directly from the graph of their integrands.
Extrema of an Accumulation Function
Since
\[ A'(x)=f(x), \]critical points of \(A\) occur where
\[ f(x)=0 \]or where \(f\) is not defined.
If \(f\) changes from positive to negative, \(A\) has a local maximum.
If \(f\) changes from negative to positive, \(A\) has a local minimum.
Reading an Accumulation Function from a Graph
If
\[ A(x)=\int_a^x f(t)\,dt, \]then the graph of \(f\) tells us several things about \(A\):
- signed area determines the value of \(A\);
- the sign of \(f\) determines whether \(A\) rises or falls;
- the magnitude of \(f\) determines how steeply \(A\) changes;
- the change in \(f\) determines the concavity of \(A\).
Why Continuity Appears in the Theorem
The standard introductory form assumes that \(f\) is continuous.
Continuity ensures that local function values accurately control local accumulation and guarantees the clean relationship
\[ \frac{d}{dx}\int_a^x f(t)\,dt=f(x). \]More advanced analysis extends these ideas under weaker hypotheses, but continuity is the natural starting point.
The Theorem Does Not Replace the Definition
A definite integral is still fundamentally defined through accumulation and Riemann-sum limits.
The Fundamental Theorem provides a powerful evaluation method when an antiderivative is available.
Thus:
\[ \boxed{ \text{Riemann sums explain what an integral is;} } \] \[ \boxed{ \text{the Fundamental Theorem explains how antiderivatives evaluate it.} } \]Common Error: Reversing the Evaluation Order
The correct rule is
\[ \boxed{ [F(x)]_a^b = F(b)-F(a) }. \]It is always
\[ \boxed{\text{upper minus lower}}, \]not lower minus upper.
Common Error: Adding \(+C\) to a Definite Integral
When evaluating
\[ \int_a^b f(x)\,dx, \]do not append \(+C\) to the final numerical result.
Any constant in the antiderivative cancels during endpoint subtraction.
Common Error: Forgetting the Chain Rule
For
\[ \frac{d}{dx} \int_a^{g(x)}f(t)\,dt, \]the answer is not merely
\[ f(g(x)). \]We must multiply by \(g'(x)\):
\[ \boxed{ f(g(x))g'(x) }. \]Common Error: Missing the Negative Sign for a Lower Limit
For
\[ \frac{d}{dx} \int_{g(x)}^b f(t)\,dt, \]the correct result is
\[ \boxed{ -f(g(x))g'(x) }. \]The negative sign comes from reversed orientation.
Common Error: Integrating When Differentiation Is Easier
For a function such as
\[ A(x)=\int_0^x e^{t^2}\,dt, \]there is no elementary antiderivative for \(e^{t^2}\).
But Part I immediately gives
\[ \boxed{A'(x)=e^{x^2}}. \]Recognize the structure before attempting unnecessary integration.
Common Error: Confusing the Variable of Integration with the Upper Limit
In
\[ \int_0^x f(t)\,dt, \]\(t\) is the integration variable and \(x\) is the variable limit.
When differentiation is complete, the integrand is evaluated at the upper limit:
\[ f(t)\longrightarrow f(x). \]A Reliable Procedure for Part I Problems
- Identify which limit depends on \(x\).
- Evaluate the integrand at that variable limit.
- Multiply by the derivative of the variable limit.
- If the variable limit is lower, attach a negative sign.
- If both limits vary, compute upper contribution minus lower contribution.
- Simplify only after the structure is correct.
A Reliable Procedure for Part II Problems
- Find an antiderivative \(F\) of the integrand.
- Write the evaluation form \([F(x)]_a^b\).
- Substitute the upper endpoint.
- Subtract the value at the lower endpoint.
- Simplify the numerical result.
- Do not add \(+C\).
- Check sign and units when the problem has an applied interpretation.
Example: Full Definite-Integral Workflow
Evaluate
\[ \int_{-1}^{2}(4x^3-2x+3)\,dx. \]An antiderivative is
\[ F(x)=x^4-x^2+3x. \]Apply the Fundamental Theorem:
\[ [x^4-x^2+3x]_{-1}^{2}. \]At the upper endpoint:
\[ F(2)=16-4+6=18. \]At the lower endpoint:
\[ F(-1)=1-1-3=-3. \]Therefore,
\[ 18-(-3)=21. \]Hence,
\[ \boxed{21}. \]Example: Full Variable-Limit Workflow
Differentiate
\[ H(x)=\int_{1-x}^{x^2+1}\sqrt{1+t^4}\,dt. \]Upper contribution:
\[ \sqrt{1+(x^2+1)^4}(2x). \]Lower contribution:
\[ - \sqrt{1+(1-x)^4}(-1). \]Therefore,
\[ \boxed{ H'(x) = 2x\sqrt{1+(x^2+1)^4} + \sqrt{1+(1-x)^4} }. \]Practice 1: Basic Accumulation Derivative
Differentiate:
\[ F(x)=\int_0^x (t^2+5)\,dt. \]Practice 2: Exponential Integrand
Differentiate:
\[ G(x)=\int_2^x e^{t^3}\,dt. \]Practice 3: Variable Upper Limit
Differentiate:
\[ H(x)=\int_0^{x^3}(1+t^2)\,dt. \]Practice 4: Variable Lower Limit
Differentiate:
\[ K(x)=\int_{x^2}^{4}\sin t\,dt. \]Practice 5: Two Variable Limits
Differentiate:
\[ P(x)=\int_x^{2x}(1+t^3)\,dt. \]Practice 6: Polynomial Definite Integral
Evaluate:
\[ \int_0^2(3x^2+2)\,dx. \]Practice 7: Trigonometric Definite Integral
Evaluate:
\[ \int_0^\pi \sin x\,dx. \]Practice 8: Exponential Definite Integral
Evaluate:
\[ \int_0^{\ln 2}e^x\,dx. \]Practice 9: Net Change
A quantity \(Q(t)\) satisfies
\[ Q'(t)=4t-1. \]Find
\[ Q(3)-Q(1). \]Practice 10: Accumulation Function Value
Let
\[ A(x)=\int_1^x 3t^2\,dt. \]Find \(A(2)\).
Practice 11: Accumulation Monotonicity
Suppose
\[ A(x)=\int_0^x f(t)\,dt \]and \(f(x)<0\) on \((2,5)\).
Is \(A\) increasing or decreasing on \((2,5)\)?
Practice 12: Accumulation Concavity
Suppose
\[ A(x)=\int_0^x f(t)\,dt \]and \(f\) is increasing on an interval.
What can you say about the concavity of \(A\)?
Practice 13: Identify the Error
A student writes
\[ \frac{d}{dx} \int_0^{x^2}\cos t\,dt = \cos(x^2). \]Identify and correct the error.
Practice 14: Upper Minus Lower
A student evaluates
\[ \int_1^3 2x\,dx \]as
\[ 1^2-3^2=-8. \]Explain the error.
Practice Answers
Practice 1
By Part I,
\[ \boxed{F'(x)=x^2+5}. \]Practice 2
\[ \boxed{G'(x)=e^{x^3}}. \]Practice 3
Evaluate the integrand at \(x^3\):
\[ 1+x^6. \]Multiply by
\[ \frac{d}{dx}(x^3)=3x^2. \]Therefore,
\[ \boxed{H'(x)=3x^2(1+x^6)}. \]Practice 4
Since the variable limit is lower,
\[ K'(x) = -\sin(x^2)(2x). \]Therefore,
\[ \boxed{K'(x)=-2x\sin(x^2)}. \]Practice 5
Upper contribution:
\[ [1+(2x)^3](2). \]Lower contribution:
\[ -(1+x^3). \]Thus,
\[ \boxed{ P'(x)=2(1+8x^3)-(1+x^3) }. \]Simplifying,
\[ \boxed{P'(x)=1+15x^3}. \]Practice 6
An antiderivative is
\[ x^3+2x. \]Therefore,
\[ [x^3+2x]_0^2 = 8+4. \]Hence,
\[ \boxed{12}. \]Practice 7
An antiderivative of \(\sin x\) is \(-\cos x\).
\[ [-\cos x]_0^\pi = -\cos\pi-(-\cos0). \] \[ =1-(-1). \]Therefore,
\[ \boxed{2}. \]Practice 8
\[ [e^x]_0^{\ln2} = e^{\ln2}-e^0. \] \[ =2-1. \]Therefore,
\[ \boxed{1}. \]Practice 9
By the net change theorem,
\[ Q(3)-Q(1) = \int_1^3(4t-1)\,dt. \]An antiderivative is
\[ 2t^2-t. \]Therefore,
\[ [2t^2-t]_1^3 = (18-3)-(2-1). \] \[ =15-1. \]Hence,
\[ \boxed{14}. \]Practice 10
\[ A(2) = \int_1^2 3t^2\,dt = [t^3]_1^2. \] \[ =8-1. \]Therefore,
\[ \boxed{7}. \]Practice 11
Since
\[ A'(x)=f(x) \]and \(f(x)<0\), we have
\[ \boxed{A\text{ is decreasing on }(2,5)}. \]Practice 12
Because
\[ A''(x)=f'(x), \]and \(f\) is increasing, \(f'(x)\) is positive where differentiable.
Therefore,
\[ \boxed{A\text{ is concave up}}. \]Practice 13
The student forgot the chain-rule factor.
Since
\[ \frac{d}{dx}(x^2)=2x, \]the correct derivative is
\[ \boxed{ 2x\cos(x^2) }. \]Practice 14
The student evaluated lower minus upper.
The correct order is upper minus lower:
\[ [x^2]_1^3 = 3^2-1^2. \]Therefore,
\[ \boxed{8}. \]What to Remember
Part I says that continuous accumulation creates an antiderivative:
\[ \boxed{ \frac{d}{dx} \int_a^x f(t)\,dt = f(x) }. \]With a variable upper limit,
\[ \boxed{ \frac{d}{dx} \int_a^{g(x)}f(t)\,dt = f(g(x))g'(x) }. \]Part II evaluates a definite integral using an antiderivative:
\[ \boxed{ \int_a^b f(x)\,dx = F(b)-F(a) }. \]The conceptual identity is
\[ \boxed{ \text{accumulated rate} = \text{net change} }. \]The Fundamental Theorem of Calculus is the bridge that turns differentiation and integration into two sides of one mathematical structure.
Where We Go Next
The Fundamental Theorem gives the core connection between derivatives and integrals. The next lesson, Substitution, develops a systematic way to reverse the chain rule when an integrand contains a function together with its derivative.
This becomes the first major general technique for finding antiderivatives beyond direct recognition.