Fundamental Theorem of Calculus (Parts 1 and 2) Explained

The Fundamental Theorem of Calculus connects differentiation and integration. It establishes them as inverse operations. This theorem bridges differential and integral calculus. Part 1 proves that an accumulation function’s derivative equals the original integrand. Also, Part 2 allows you to evaluate definite integrals. You do this by evaluating an antiderivative at the interval’s endpoints.

What is Part 1 of the Fundamental Theorem of Calculus?

Part 1 of the theorem deals with accumulation functions. These functions measure the area accumulated under a curve. This area goes from a fixed starting point a up to a variable boundary x. Formally, if f is continuous on the closed interval [a, b], then you define the function F by:

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

is continuous on [a, b]. It is also differentiable on the open interval (a, b). Its derivative is simply:

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

Geometric and physical intuition

Geometrically, F(x) shows the net area. This area is under the curve y = f(t) from t = a to t = x. You can shift the right boundary x to the right by a tiny distance h. As a result, the accumulated area increases. It increases by a thin rectangular strip. This strip has width h. Its height is approximately equal to f(x). So, the change in area is roughly f(x) · h. This means the instantaneous rate of change of area with respect to x is f(x).

A physical example makes this concept intuitive. Imagine water flowing into a reservoir. Its rate is f(t) liters per minute. Let F(x) be the total water in the reservoir at time x. The total volume increases at time x. This increase equals the current inflow rate, f(x). Therefore, differentiating the total accumulated amount gives the rate of accumulation, a concept key in the fundamental theorem of calculus.

Notice the variable t inside the integrand. In this context, t is a dummy variable of integration. It ranges from a to x. Also, it disappears once you evaluate the definite integral. Using t inside the integral prevents confusion for the fundamental theorem of calculus. This is because it differs from the upper limit x, which is the independent variable of F(x).

How to differentiate accumulation functions step-by-step

You can differentiate integral expressions directly. Apply Part 1 of the fundamental theorem of calculus or combine it with the chain rule. This means you do not need to calculate the integral itself. This is vital when an integrand lacks an elementary antiderivative.

1. Standard variable upper limit: If the upper limit is x and the lower limit is a constant, replace the dummy variable with x.

   $$\frac{d}{dx}\int_0^x \cos(t^2)\,dt = \cos(x^2)$$

   Because \(\int \cos(t^2)\,dt\) cannot be expressed with elementary functions, Part 1 provides the derivative directly.

2. Upper limit as a function of x (Chain Rule): If the upper limit is a differentiable function g(x), substitute g(x) into the integrand and multiply by g‘(x):

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

   For example:

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

3. Variable lower limit: If the variable appears in the lower limit, reverse the integration limits by multiplying by -1:

   $$\frac{d}{dx}\int_x^5 e^{t^2}\,dt = \frac{d}{dx}\left(-\int_5^x e^{t^2}\,dt\right) = -e^{x^2}$$

4. Both limits variable: When both limits depend on x, split the integral at any constant c within the domain using integral properties:

   $$\int_{h(x)}^{g(x)} f(t)\,dt = \int_c^{g(x)} f(t)\,dt - \int_c^{h(x)} f(t)\,dt$$

   Differentiating both pieces yields:

   $$\frac{d}{dx}\int_{h(x)}^{g(x)} f(t)\,dt = f\big(g(x)\big) \cdot g'(x) - f\big(h(x)\big) \cdot h'(x)$$

   For example, differentiating \(\int_x^{x^2} t^3\,dt\):

   $$\frac{d}{dx}\int_x^{x^2} t^3\,dt = (x^2)^3 \cdot 2x - x^3 \cdot 1 = 2x^7 - x^3$$

   You can verify this result directly: integrating \(t^3\) gives \([t^4/4]_x^{x^2} = x^8/4 - x^4/4\), and taking its derivative gives \(2x^7 - x^3\).

5. Finding critical points of an integral function: Calculus exams often ask you to analyze functions defined by integrals. For instance, you can locate local extrema of \(F(x) = \int_0^x (t^2 - 4)\,dt\). First, find where \(F'(x) = 0\). By Part 1, \(F'(x) = x^2 - 4\). Setting \(F'(x) = 0\) gives \(x = \pm 2\). Apply standard critical point analysis. \(F'\) changes from positive to negative at \(x = -2\). This indicates a local maximum. Also, \(F'\) changes from negative to positive at \(x = 2\). This indicates a local minimum. You never need to compute the integral itself.

How do you prove FTC Part 1?

To prove Part 1, set up the derivative of \(F(x)\) using the limit definition of the derivative:

$$F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}$$

Using the definition \(F(x) = \int_a^x f(t)\,dt\), the numerator represents the difference between two integrals:

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

Substituting this back into the difference quotient gives:

$$\frac{F(x+h) - F(x)}{h} = \frac{1}{h} \int_x^{x+h} f(t)\,dt$$

The term \(\frac{1}{h} \int_x^{x+h} f(t)\,dt\) represents the average value of \(f(t)\). This value is on the interval $[x, x+h]$. Because \(f\) is continuous on $[x, x+h]$, the Extreme Value Theorem, a fundamental theorem of calculus, guarantees something. It guarantees that \(f\) attains a minimum value \(m_h\) and a maximum value \(M_h\) on this interval. Therefore:

$$m_h \cdot h \le \int_x^{x+h} f(t)\,dt \le M_h \cdot h$$

Dividing all parts of the inequality by \(h\) yields:

$$m_h \le \frac{1}{h} \int_x^{x+h} f(t)\,dt \le M_h$$

As \(h \to 0\), the interval $[x, x+h]$ shrinks to the single point \(x\). Because \(f\) is continuous, both \(m_h \to f(x)\) and \(M_h \to f(x)\). By the squeeze theorem, the middle expression must also approach \(f(x)\):

$$F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = f(x)$$

This completes the proof of Part 1.

What is Part 2 of the Fundamental Theorem of Calculus?

Part 2, often called the Evaluation Theorem, provides a practical method for evaluating definite integrals without taking limits of Riemann sums. This fundamental theorem of calculus states that if \(f\) is continuous on $[a, b]$ and \(F\) is any antiderivative of \(f\) on $[a, b]$ (meaning \(F'(x) = f(x)\)), then:

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

You can write \(F(b) - F(a)\) in two ways. Use the evaluation bracket notation \([F(x)]_a^b\). Also, you can use \(F(x)\big|_a^b\).

Because any two antiderivatives of \(f\) differ only by a constant \(C\), the arbitrary constant cancels out during evaluation:

$$\big(F(b) + C\big) - \big(F(a) + C\big) = F(b) - F(a)$$

Consequently, you can omit \(C\) when computing definite integrals.

Step-by-step worked examples for FTC Part 2

1. Polynomial integrand: Evaluate \(\int_1^2 3x^2\,dx\).

   Find an antiderivative of \(3x^2\), which is \(x^3\). Evaluate from 1 to 2:

   $$\int_1^2 3x^2\,dx = \Big[x^3\Big]_1^2 = (2)^3 - (1)^3 = 8 - 1 = 7$$

2. Trigonometric integrand: Evaluate \(\int_0^{\pi/2} \cos(x)\,dx\).

   An antiderivative of \(\cos(x)\) is \(\sin(x)\):

   $$\int_0^{\pi/2} \cos(x)\,dx = \Big[\sin(x)\Big]_0^{\pi/2} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1$$

3. Logarithmic/Rational integrand: Evaluate \(\int_1^{e^2} \frac{1}{x}\,dx\).

   An antiderivative of \(1/x\) for $x > 0$ is \(\ln(x)\):

   $$\int_1^{e^2} \frac{1}{x}\,dx = \Big[\ln(x)\Big]_1^{e^2} = \ln(e^2) - \ln(1) = 2 - 0 = 2$$

4. Power integrand with fractional exponents: Evaluate \(\int_1^4 \sqrt{x}\,dx\).

   Rewrite \(\sqrt{x}\) as \(x^{1/2}\). Using the derivative rule in reverse, the antiderivative is \(\frac{2}{3}x^{3/2}\):

   $$\int_1^4 x^{1/2}\,dx = \left[\frac{2}{3}x^{3/2}\right]_1^4 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{14}{3}$$

5. Signed area and zero net integrals: Evaluate \(\int_0^1 (4x^3 - 2x)\,dx\).

   An antiderivative is \(x^4 - x^2\):

   $$\int_0^1 (4x^3 - 2x)\,dx = \Big[x^4 - x^2\Big]_0^1 = (1 - 1) - (0 - 0) = 0$$

   A result of zero does not mean the region has no area. A definite integral calculates signed area—the region above the horizontal axis counts as positive area, while the region below counts as negative area. Here, the positive area exactly equals the magnitude of the negative area.

How do you prove FTC Part 2?

The proof of Part 2 relies directly on Part 1 and the Mean Value Theorem.

Define an accumulation function \(G(x) = \int_a^x f(t)\,dt\). By Part 1, \(G(x)\) is an antiderivative of \(f(x)\), meaning \(G'(x) = f(x)\).

Suppose \(F(x)\) is any other antiderivative of \(f(x)\) on $[a, b]$, so \(F'(x) = f(x)\). By a standard corollary of the Mean Value Theorem, two functions with identical derivatives on an interval differ by a constant \(C\), which is a concept central to the fundamental theorem of calculus:

$$F(x) = G(x) + C$$

Now evaluate \(F(b) - F(a)\):

$$F(b) - F(a) = \big(G(b) + C\big) - \big(G(a) + C\big) = G(b) - G(a)$$

Substitute the definition of \(G(x)\) into \(G(b)\) and \(G(a)\):

$$G(b) = \int_a^b f(t)\,dt$$

$$G(a) = \int_a^a f(t)\,dt = 0$$

Therefore:

$$F(b) - F(a) = \int_a^b f(t)\,dt - 0 = \int_a^b f(x)\,dx$$

This completes the proof of Part 2.

When does the Fundamental Theorem of Calculus fail?

The fundamental theorem of calculus has a crucial hypothesis in both parts. The function \(f(x)\) must be continuous on the entire closed interval $[a, b]$. For example, if you apply Part 2 across an interior point of discontinuity, you will get incorrect or nonsense results.

Consider the classical paradox:

$$\int_{-1}^1 \frac{1}{x^2}\,dx$$

If you naively apply Part 2 using the antiderivative \(F(x) = -1/x\), you get:

$$\Big[-\frac{1}{x}\Big]_{-1}^1 = \left(-\frac{1}{1}\right) - \left(-\frac{1}{-1}\right) = -1 - 1 = -2$$

This answer is clearly impossible: \(1/x^2 > 0\) for all real \(x \ne 0\), so the area under its graph cannot be negative. The FTC fails here because \(f(x) = 1/x^2\) has an infinite discontinuity at \(x = 0\), which lies inside $[-1, 1]$. This is actually an improper integral that diverges to \(+\infty\).

If a function has jump discontinuities or removable discontinuities, you cannot apply FTC across the whole interval at once. Instead, split the interval into continuous sub-intervals and evaluate each piece separately.

How does the Net Change Theorem apply FTC to real problems?

Part 2 is often called the Net Change Theorem. It states that the integral of a rate of change is the net change for a fundamental theorem of calculus. This net change, however, occurs in the original quantity over time.

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

Common applications include:

• Kinematics: Integrating velocity \(v(t) = s'(t)\) gives net displacement \(s(b) - s(a)\).
• Fluid dynamics: You can integrate the volumetric flow rate \(r(t) = V'(t)\). As a result, this gives you the total change in fluid volume \(V(b) - V(a)\).
• Economics: Integrating marginal cost \(C'(x)\) gives the total change in production cost \(C(b) - C(a)\).

Worked example: Displacement vs. Total Distance Traveled

A particle moves along a line with velocity \(v(t) = 3t^2 - 12\) meters per second for \(0 \le t \le 3\).

1. Net Displacement: Integrate velocity directly over $[0, 3]$:

$$\text{Displacement} = \int_0^3 (3t^2 - 12)\,dt = \Big[t^3 - 12t\Big]_0^3 = (27 - 36) - (0 - 0) = -9 \text{ meters}$$

The particle ends up 9 meters to the left of its starting position.

2. Total Distance Traveled: Distance requires integrating speed $|v(t)|$. First, find where velocity changes sign by setting \(v(t) = 0\):

$$3t^2 - 12 = 0 \implies t^2 = 4 \implies t = 2 \text{ seconds}$$

On $[0, 2]$, \(v(t) \le 0\) (moving left). On $[2, 3]$, \(v(t) \ge 0\) (moving right). Therefore, split the integral at \(t = 2\) and take absolute values:

$$\int_0^2 (3t^2 - 12)\,dt = \Big[t^3 - 12t\Big]_0^2 = (8 - 24) = -16 \text{ meters (16 meters left)}$$

$$\int_2^3 (3t^2 - 12)\,dt = \Big[t^3 - 12t\Big]_2^3 = (27 - 36) - (8 - 24) = -9 - (-16) = 7 \text{ meters (7 meters right)}$$

$$\text{Total Distance} = |-16| + |7| = 16 + 7 = 23 \text{ meters}$$

How do Part 1 and Part 2 compare?

While both parts express the inverse relationship between derivatives and integrals, they serve distinct operational goals:

Common mistakes to avoid when using the FTC

• Forgetting the chain rule in Part 1: You might write \(\frac{d}{dx}\int_0^{x^2} f(t)\,dt = f(x^2)\). However, this is incorrect. Instead, you need \(f(x^2) \cdot 2x\). Always multiply by the derivative of the upper boundary.
• Subtracting \(f(a)\) when differentiating in Part 1: The constant lower limit contributes zero. This is because \(\frac{d}{dx}\int_a^c f(t)dt = 0\).
• Reversing the subtraction order in Part 2: Always calculate \(F(\text{upper limit}) - F(\text{lower limit})\). Reversing them, however, flips the sign of your answer.
• Ignoring discontinuities inside the integration interval: Do not blindly compute antiderivatives across asymptotic points. This leads to incorrect answers. For example, you might get negative areas for positive functions.
• Confusing net displacement with total distance: Displacement allows positive and negative areas to cancel out. However, distance requires integrating $|v(t)|$. Therefore, you must split the integral at turning points.

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

Test your understanding with these practice problems before checking the answers.

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

Answers and solutions:

1. \(\sqrt{1 + x^3}\) — Direct application of Part 1. Replace \(t\) with \(x\).
2. \(\sin^2(x)\cos(x)\) Apply Part 1 with the chain rule. First, substitute \(\sin(x)\) for \(t\). This gives you \(\sin^2(x)\). Then, multiply by \(\frac{d}{dx}\sin(x) = \cos(x)\).
3. \(2\) — An antiderivative of \(\sin(x)\) is \(-\cos(x)\). First, evaluate \([-\cos(x)]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) + 1 = 2\).
4. \(3x^5 - 2x^3\) — You can split this or apply the two-boundary formula. First, use the formula: \((x^3) \cdot \frac{d}{dx}(x^3) - (x^2) \cdot \frac{d}{dx}(x^2)\). This gives you \(x^3(3x^2) - x^2(2x)\). As a result, you get \(3x^5 - 2x^3\).
5. \(9\) — An antiderivative of \(2x\) is \(x^2\). Evaluate \([x^2]_0^3 = 3^2 - 0^2 = 9\).

Frequently asked questions

What is the main difference between Part 1 and Part 2 of the FTC?

Part 1 shows that differentiation and integration are inverse processes by taking the derivative of an accumulation function. Part 2, however, gives a direct computational rule for finding the value of a definite integral using an antiderivative.

Does every continuous function have an antiderivative?

Yes, Part 1 of the fundamental theorem of calculus explicitly proves this. If \(f(x)\) is continuous on $[a, b]$, then \(F(x) = \int_a^x f(t)\,dt\) is an antiderivative of \(f(x)\). Even functions like \(e^{-x^2}\) have antiderivatives. For example, this is true for the Gaussian integral. You cannot write them in terms of elementary algebraic functions.

Why do we use a dummy variable inside an accumulation integral?

The dummy variable (such as \(t\)) acts as a placeholder. It gets integrated out inside the integrand. Using \(t\) inside \(\int_a^x f(t)\,dt\) avoids confusion for the fundamental theorem of calculus. It prevents confusing the variable of integration with the upper limit \(x\). Also, \(x\) defines the input variable for the overall accumulation function.

What does the Fundamental Theorem of Calculus represent geometrically?

Geometrically, Part 1 of the fundamental theorem of calculus states the rate of change of the area under a curve. This rate equals the height of the curve at that boundary. This happens as the right boundary moves. Part 2 states the total accumulated area under a curve between \(a\) and \(b\). As a result, this area equals the net change in its antiderivative function over that interval.

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.