Fundamental Theorem of Calculus (Parts 1 and 2) Explained

The Fundamental Theorem of Calculus (FTC) says that differentiation and integration are inverse operations. It is the reason we can compute areas with antiderivatives instead of adding up infinitely many rectangles.

The theorem has two parts. Part 1 says that if you build a function by accumulating area, its derivative is the original function. Part 2 says that to find a definite integral, you only need an antiderivative. Together they connect the two big ideas of the course: the derivative (a rate) and the integral (an accumulated total).

Part 1: the derivative of an accumulation function

If \( f \) is continuous on \( [a, b] \) and

$$F(x) = \int_a^x f(t)\,dt$$

then \( F \) is differentiable and

$$F'(x) = f(x)$$

Intuition: \( F(x) \) is the area under \( f \) from \( a \) up to \( x \). Nudge \( x \) forward by a tiny \( h \): the area grows by a thin strip of width \( h \) and height about \( f(x) \). So the rate of change of area is the height of the curve.

A physical picture helps too. Imagine water flowing into a tank at a rate of \( f(t) \) liters per minute, and let \( F(x) \) be the total amount that has flowed in by time \( x \). How fast is the total growing right now? Exactly at the current flow rate, \( f(x) \). That sentence is Part 1.

The letter \( t \) inside the integral is a dummy variable. It runs from \( a \) to \( x \), and it disappears once the integral is evaluated, leaving a function of \( x \) alone. Using a different letter keeps the upper limit and the variable of integration from being confused.

Examples

1. \( \frac{d}{dx}\int_0^x\cos(t^2)\,dt = \cos(x^2) \). No integration needed (and this integral has no elementary formula anyway).
2. With the chain rule: if the upper limit is \( g(x) \),

$$\frac{d}{dx}\int_a^{g(x)}f(t)\,dt = f\big(g(x)\big)\,g'(x)$$

So \( \frac{d}{dx}\int_1^{x^2}\ln t\,dt = \ln(x^2)\cdot 2x \).

1. Variable lower limit: flip the limits and add a minus sign. \( \frac{d}{dx}\int_x^5 e^{t^2}dt = -e^{x^2} \).

2. Both limits variable. Split at any constant, then apply the rule to each piece. For \( \frac{d}{dx}\int_x^{x^2}t^3\,dt \), the upper limit contributes \( (x^2)^3\cdot2x \) and the lower limit contributes \( -x^3 \):

$$\frac{d}{dx}\int_x^{x^2}t^3\,dt = 2x^7 - x^3$$

You can confirm this one directly: the integral equals \( \frac{x^8}{4} - \frac{x^4}{4} \), and its derivative is \( 2x^7 - x^3 \).

1. An exam-style question. Where does \( F(x) = \int_0^x(t^2 - 4)\,dt \) have a local maximum? By Part 1, \( F'(x) = x^2 - 4 \), which is zero at \( x = \pm2 \). The sign of \( F' \) goes from positive to negative at \( x = -2 \), so that is a local maximum, and from negative to positive at \( x = 2 \), a local minimum. You never had to compute \( F \) itself; this is the same critical point analysis you use for any function.

Proof of Part 1

Form the difference quotient. Since the integral from \( a \) to \( x + h \) minus the integral from \( a \) to \( x \) is the integral over the thin strip,

$$\frac{F(x+h) - F(x)}{h} = \frac1h\int_x^{x+h}f(t)\,dt$$

The right side is the average value of \( f \) on the small interval from \( x \) to \( x + h \). Because \( f \) is continuous, it has a minimum \( m_h \) and a maximum \( M_h \) on that interval, and the average lies between them. As \( h \to 0 \), the interval shrinks to the single point \( x \), so both \( m_h \) and \( M_h \) approach \( f(x) \). By the squeeze theorem, the difference quotient approaches \( f(x) \) too, which says \( F'(x) = f(x) \).

Part 2: evaluating definite integrals

If \( F \) is any antiderivative of \( f \) on \( [a,b] \), then

$$\int_a^b f(x)\,dx = F(b) - F(a)$$

Examples

1. \( \int_1^2 3x^2\,dx = \big[x^3\big]_1^2 = 8 - 1 = 7 \)
2. \( \int_0^{\pi/2}\cos x\,dx = \big[\sin x\big]_0^{\pi/2} = 1 \)
3. \( \int_1^{e^2}\frac{dx}{x} = \big[\ln x\big]_1^{e^2} = 2 \)
4. \( \int_1^4\sqrt x\,dx = \left[\frac23x^{3/2}\right]_1^4 = \frac23(8 - 1) = \frac{14}{3} \)
5. \( \int_0^1(4x^3 - 2x)\,dx = \big[x^4 - x^2\big]_0^1 = 0 \). A zero answer does not mean there is no area: the region below the axis exactly cancels the region above it. A definite integral measures signed area.

The constant \( C \) cancels in \( F(b) - F(a) \), so you never need it here.

Proof of Part 2

Let \( G(x) = \int_a^x f(t)\,dt \). By Part 1, \( G \) is an antiderivative of \( f \). Any two antiderivatives on an interval differ by a constant (a consequence of the Mean Value Theorem), so \( F(x) = G(x) + C \) for some \( C \). Then

$$\begin{aligned} F(b) - F(a) &= \big(G(b) + C\big) - \big(G(a) + C\big) \\ &= G(b) - G(a) \\ &= \int_a^b f(t)\,dt - 0 \end{aligned}$$

since \( G(a) \) is an integral over an interval of zero width. That is exactly Part 2.

The fine print: continuity matters

Part 2 needs \( f \) to be continuous on the whole interval. \( \int_{-1}^1\frac{dx}{x^2} \) looks like \( \left[-\frac1x\right]_{-1}^1 = -2 \), which is impossible for a positive function. The problem is the discontinuity at 0; this is really a divergent improper integral.

Net change theorem

Part 2 has a practical reading: integrating a rate gives total change.

$$\int_a^b v(t)\,dt = s(b) - s(a)$$

Velocity integrates to displacement, flow rate to volume, marginal cost to total cost.

Worked example. A particle moves with velocity \( v(t) = 3t^2 - 12 \) meters per second for \( 0 \le t \le 3 \). Its displacement is

$$\int_0^3(3t^2 - 12)\,dt = \big[t^3 - 12t\big]_0^3 = 27 - 36 = -9$$

so it ends 9 meters to the left of where it started. The total distance traveled is different, because the particle turns around at \( t = 2 \), where \( v = 0 \). Integrating the speed \( |v(t)| \) in two pieces gives 16 meters moving left and 7 meters moving right, for a total of 23 meters.

Why “fundamental”?

Before the FTC, areas and tangent lines were two separate problems. The theorem shows they are two sides of one idea, which is why every derivative rule instantly gives an integration rule.

It also explains why Riemann sums and antiderivatives give the same answer. A Riemann sum approximates area by rectangles; Part 2 says that the limit of those sums can be read off from any antiderivative. Isaac Newton and Gottfried Leibniz each developed this connection independently in the late 1600s, and it is what turned calculus into a practical tool.

Common mistakes

• Forgetting the chain rule in Part 1. \( \frac{d}{dx}\int_0^{x^2}f(t)\,dt \) is \( f(x^2)\cdot2x \), not \( f(x^2) \).
• Subtracting \( f(a) \) in Part 1. The derivative of \( \int_a^x f(t)\,dt \) is just \( f(x) \). The constant lower limit contributes nothing.
• Subtracting in the wrong order in Part 2. It is always upper limit minus lower limit, \( F(b) - F(a) \). Put \( F(a) \) in parentheses so its sign is handled correctly.
• Ignoring discontinuities. Check that the integrand is continuous on the whole interval before applying Part 2.
• Confusing displacement with distance. To get total distance, integrate \( |v(t)| \), which means splitting at every point where \( v \) changes sign.

Try it: definite integrals with working

The solver finds an antiderivative, then applies Part 2:

For more problems, the definite integral calculator shows each step, including the substitution of the limits.

Practice problems

Try these before checking the answers.

1. \( \frac{d}{dx}\int_0^{x}\sqrt{1 + t^3}\,dt \)
2. \( \frac{d}{dx}\int_0^{\sin x}t^2\,dt \)
3. \( \int_0^{\pi}\sin x\,dx \)
4. \( \frac{d}{dx}\int_{x^2}^{x^3}t\,dt \)
5. \( \int_0^3 2x\,dx \)

Answers: (1) \( \sqrt{1 + x^3} \); (2) \( \sin^2x\cos x \); (3) \( 2 \); (4) \( 3x^5 - 2x^3 \); (5) \( 9 \).

For (4), the upper limit gives \( x^3\cdot3x^2 \) and the lower limit gives \( -x^2\cdot2x \).

FAQ

What’s the difference between Part 1 and Part 2?

Part 1: differentiating an integral gives back the integrand. Part 2: to evaluate a definite integral, subtract antiderivative values.

Does every continuous function have an antiderivative?

Yes. Part 1 constructs one: \( \int_a^x f(t)\,dt \). It just may not have an elementary formula, as with \( e^{-x^2} \) (Gaussian integral).

Why do we use t instead of x inside the integral?

Because \( x \) is already being used as the upper limit. The variable of integration is a placeholder that disappears after integrating, so any letter other than \( x \) avoids confusion.

What does the theorem mean geometrically?

The rate at which the area under a curve grows, as you move the right edge, equals the height of the curve at that edge. Adding up those growth rates from \( a \) to \( b \) gives back the total area.

Further reading

• Computing Definite Integrals (Paul’s Online Math Notes) — Part 2 in action on many examples, including piecewise and absolute value integrands.
• Definition of the Definite Integral (Paul’s Online Math Notes) — the limit definition, integral properties and Part 1 with chain rule examples.