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Fundamental Theorem of Calculus

The fundamental theorem of calculus (FTC) tells us the connection between differentiation and integration. This connection is discovered by Sir Isaac Newton and Gottfried Wilhelm Leibniz during late 1600s. There are two parts of the FTC: FTC 1 and FTC 2

We are aware of the fact that differentiation and integration are the reverse processes of each other and the first FTC justifies it. We are also aware that a definite integral is evaluated first by evaluating the indefinite integral and then substituting the upper and lower bounds, and this process is justified by the second FTC.

1.	What is the Fundamental Theorem of Calculus?
2.	First Fundamental Theorem of Calculus (Part 1)
3.	Second Fundamental Theorem of Calculus (Part 2)
4.	Applications of Fundamental Theorem of Calculus
5.	FAQs on Fundamental Theorem of Calculus

What is the Fundamental Theorem of Calculus?

There are two parts of the fundamental theorem of calculus. These theorems are powerful as they are helpful in evaluating the definite integral (or they are helpful in calculating the area between the curves) without using the Riemann sums. Here are the statements of the fundamental theorems of calculus.

Fundamental Theorem of Calculus Formula

The fundamental theorem of calculus has two formulas:

The part 1 (FTC 1) is d/dx ∫_a^x f(t) dt = f(x)
The part 2 (FTC 2) is ∫_a^b f(t) dt = F(b) - F(a), where F(x) = ∫_a^b f(x) dx

Let us learn in detail about each of these theorems along with their proofs.

First Fundamental Theorem of Calculus (Part 1)

The first fundamental theorem of calculus (FTC Part 1) is used to find the derivative of an integral and so it defines the connection between the derivative and the integral. Using this theorem, we can evaluate the derivative of a definite integral without actually evaluating the definite integral. The first fundamental theorem of calculus (FTC 1) is stated as follows.

"If f(x) is a function that is continuous over [a, b] and differentiable over (a, b) and if F(x) is defined as F(x) = ∫_a^x f(t) dt then F'(x) = f(x) over the interval [a, b]" (OR)
"d/dx ∫_a^x f(t) dt = f(x)"

Let us prove this theorem now.

Proof

By the definition of the derivative of a function,

F'(x) = lim_h _→ ₀[F(x+h)-F(x)] / h

It is given that F(x) = ∫_a^x f(t) dt. Using this definition in the above equation,

F'(x) = lim_h _→ ₀ (1/h) [∫_a^x+h f(t) dt - ∫_a^x f(t) dt]

By a property of definite integrals, ∫_a^b f(x) dx = - ∫_b^a f(x) dx. Using this in the above equation,

F'(x) = lim_h _→ ₀ (1/h) [∫_a^x+h f(t) dt + ∫_x^a f(t) dt]

By another property of definite integrals, ∫_a^b f(x) dx + ∫_b^c f(x) dx = ∫_a^c f(x) dx. Using this in the above equation,

F'(x) = lim_h _→ 0 (1/h) ∫_x^x+h f(t) d t ... (1)

Since f(x) is continuous on [x, x + h] (this is because f(x) is continuous on [a, b] and [x, x + h] is a subinterval of [a, b]), by mean value theorem, there exists at least one point c in the interval [x, x + h] such that,

f(c) = (1/(x+h-x) ∫_x^x+h f(x) d x

(or) f(c) = (1/h) ∫_x^x+h f(x) d x

(Recalling mean value theorem: If f(x) is continuous on [a, b], then there exists atleast some point c in [a, b] such that f(c)=[1/(b-a)] ∫_a^b f(x) dx)

Substituting this in (1), we get

F'(x) = lim_{h → 0} f(c) ... (2)

Since f(x) is continuous on [x, x + h] and since c is also present in this interval, by the definition of continuity,

lim_{h → 0} f(c) = f(x)

Substituting this in (2), we get

F'(x) = f(x)

Hence the first fundamental theorem of calculus is proved.

Second Fundamental Theorem of Calculus (Part 2)

The second fundamental theorem of calculus (FTC Part 2) says the value of a definite integral of a function is obtained by substituting the upper and lower bounds in the antiderivative of the function and subtracting the results in order. Usually, to calculate a definite integral of a function, we will divide the area under the graph of that function lying within the given interval into many rectangles and then we add the areas of all such rectangles (this process is named as Riemann integration). This theorem helps to evaluate a definite integral without using the Riemann sum (or calculating the area under the curves). The second fundamental theorem of calculus (FTC 2) is stated as follows.

"If f(x) is a continuous function over [a, b] and if F(x) is some antiderivative of f(x) (i.e., F'(x) = f(x)) then ∫_a^b f(x) dx = F(b) - F(a)"

Let us prove this theorem now.

Proof

It is given that F(x) is an anti derivative of f(x). i.e.,

F'(x) = f(x) ... (1)

Let us define a new function g(x) such that

g(x) = ∫_a^x f(t) dt.

Then by the first part of the fundamental theorem of calculus (FTC 1), g'(x) = f(x) ... (2)

Let us define another function h(x) such that

h(x) = g(x) - F(x), where x is in [a, b]

Differentiating on both sides,

h'(x) = g'(x) - F'(x)

= f(x) - f(x) (From (1) and (2))

= 0

We know that h(x) is continuous on [a, b] (as both g(x) and F(x) are continuous on the same interval) and from the above equation h'(x) = 0. Thus, h(x) is a constant function over [a, b] and hence

h(b) = h(a)

By the definition of h(x),

g(b) - F(b) = g(a) - F(a)

By the definition of g(x),

∫_a^b f(t) dt - F(b) = ∫_a^a f(t) dt - F(a)

By a property of definite integrals, ∫_a^a f(t) dt = 0. Thus, the above equation becomes

∫_a^b f(t) dt - F(b) = - F(a)

(or) ∫_a^b f(t) dt = F(b) - F(a)

Hence the second fundamental theorem of integral calculus is proved.

Applications of Fundamental Theorem of Calculus

The fundamental theorem of calculus gives a very strong relation between derivative and integral.
It is helpful to evaluate a definite integral without using Riemann sum.
It is used to find the area under a curve easily.
It is used to find the derivative of an integral.

Important Notes on Fundamental Theorem of Calculus:

Using FTC 1, d/dx ∫_a^x f(t) dt = f(x), where 'a' is a constant.
Using FTC 2, to evaluate the integral ∫_a^b f(t) dt, we will first evaluate the indefinite integral ∫ f(t) dt = F(t), substitute the upper bound and lower bound, and then subtract them.
i.e., ∫_a^b f(t) dt = F(b) - F(a).
FTC 1 is used to find the derivative of an integral whereas FTC 2 is used to evaluate a definite integral.
If ∫ f(t) dt = F(t), then ∫_a^b f(t) dt is F(t)|_a^b = F(b) - F(a).

☛Related Topics:

Fundamental Theorem of Calculus Examples

Example 1: Find d/dx ∫₃^x (3+t) / (1+t³) d t using the fundamental theorem of calculus.

Solution:

The first part of the fundamental theorem of calculus (FTC 1) says, d/dx ∫_a^x f(t) dt = f(x).

We can apply this for our problem as the lower bound of the given integral is a constant (which is 3) and the upper bound is a variable (which is x). Then we get

d/dx ∫₃^x (3+t) / (1+t³) d t = (3+x) / (1+x³)

Answer: d/dx ∫₃^x (3+t) / (1+t³) d t = (3+x) / (1+x³).
Example 2: Evaluate the following derivative of the integral: (d/dx) ∫_x^2x cos t² dt.

Solution:

Let us recall the first part of the fundamental theorem of calculus (FTC 1) which says d/dx ∫_a^x f(t) dt = f(x).

Using the properties of definite integrals, we can write the given integral as follows.

∫_x^2x cos t² dt = ∫_x⁰ cos t² dt + ∫₀^2x cos t² dt

= - ∫₀^x cos t² dt + ∫₀^2x cos t² dt

Now we will take the derivative on both sides.

d/dx ∫_x^2x cos t² dt = - (d/dx) ∫₀^x cos t² dt + (d/dx) ∫₀^2x cos t² dt ... (1)

The value of the first derivative directly follows from FTC 1. i.e., (d/dx) ∫₀^x cos t² dt = cos x².

To evaluate the second derivative (which is (d/dx) ∫₀^2x, let us assume that u = 2x. Then

(d/dx) ∫₀^2x cos t² dt = d/dx ∫₀^u cos t² dt

= (d/du) ∫₀^u cos t² dt · (du/dx) (By chain rule)

= cos u² · (2) (By FTC 1)

= 2 cos u²

= 2 cos (4x²) (∵ u = 2x)

Substituting the values in (1),

(d/dx) ∫_x^2x cos t² dt = - cos x² + 2 cos (4x²)

Answer: (d/dx) ∫_x^2x cos t² dt = 2 cos (4x²) - cos x².
Example 3: Find the value of the integral ∫_-1³ (x² - 3) dx using the fundamental theorem of calculus.

Solution:

Using FTC 2, ∫_a^b f(x) dx = F(b) - F(a) where F(x) = ∫ f(x) dx (or) F'(x) = f(x).

So first we will evaluate the indefinite integral ∫ (x² - 3) dx.

∫ (x² - 3) dx = ∫ (x²)dx - ∫ 3 dx

= x³/3 - 3x + C (∵ ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C)

= F(x)

BY FTC 2,

∫_-1³ (x² - 3) dx = x³/3 - 3x + C |_-1³

= [3³/3 - 3(3)] - [(-1)³/3 - 3(-1) ]

= 9 - 9 + 1/3 - 3

= -8/3

Answer: ∫_-1³ (x² - 3) dx = -8/3.

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Practice Questions on Fundamental Theorem of Calculus

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FAQs on Fundamental Theorem of Calculus

What are Two Parts of the Fundamental Theorem of Calculus?

The fundamental theorem of calculus (FTC) tells us the relationship between derivatives and integrals. There are two parts of FTC.

FTC 1: d/dx ∫_a^x f(t) dt = f(x).
FTC 2: ∫_a^b f(x) dx = F(b) - F(a) where F(x) = ∫ f(x) dx

What is the Fundamental Theorem of Calculus Part 1 Formula?

Part 1 of the fundamental theorem of calculus is used to differentiate an integral. It says d/dx ∫_a^x f(t) dt = f(x). Note that to apply this theorem, the lower bound of the integral must be a constant.

What Is the Fundamental Theorem of Calculus Part 2 Formula?

The part 2 of fundamental theorem of calculus is used to evaluate a definite integral ∫_a^b f(x) dx by first evaluating the indefinite integral ∫ f(x) dx = F(x) and then finding the difference of the values of F(x) at the bounds of the integral. i.e., ∫_a^b f(x) dx = F(b) - F(a).

What Are the Other Names of Second Fundamental Theorem of Integral Calculus?

The second fundamental theorem of calculus says ∫_a^b f(x) dx = F(b) - F(a) where F'(x) = f(x) and is also known as "evaluation theorem" (or) " Newton–Leibniz theorem".

Who Discovered Fundamental Theorem of Calculus?

The fundamental theorem of calculus is discovered by Sir Isaac Newton and Gottfried Wilhelm Leibniz during the late 1600s and early 1700s.

What is FTC Calculus?

FTC in Calculus stands for "fundamental theorem of calculus". It has two parts in it.

1st FTC says d/dx ∫_a^x f(t) dt = f(x).
2nd FTC says ∫_a^b f(x) dx = F(b) - F(a) where F(x) = ∫ f(x) dx

How Many Types of Fundamental Theorem of Integral Calculus Are There?

There are two types of the fundamental theorems of calculus. They are:

Type 1: d/dx ∫_a^x f(t) dt = f(x).
Type 2: ∫_a^b f(x) dx = F(b) - F(a) where F'(x) = f(x) (or) F(x) = ∫ f(x) dx.

How To Find the Derivative of an Integral Using the Fundamental Theorem of Calculus?

The derivative of an integral can be found by using the first part fundamental theorem of calculus which says d/dx ∫_a^x f(t) dt = f(x), where 'a' is a constant.

How To Evaluate a Definite Integral Using the Fundamental Theorem of Calculus?

A definite integral can be evaluated using the second fundamental theorem of calculus which says ∫_a^b f(x) dx = F(b) - F(a) where F(x) is the value of the indefinite integral ∫ f(x) dx.

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