Calculus is the mathematical study of continuous change. It has two main branches – differential calculus (concerning rates of change and slopes of curves) and integral calculus (concerning the accumulation of quantities and the areas under and between curves). The Fundamental theorem of calculus links these two branches. In this article, we will look at the two fundamental theorems of calculus and understand them with the help of some examples.

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## Area Function

Take a look at the following image

By definition, ∫_{a}^{b} f(x) dx is the area of the region bounded by the curve y = f(x), the x-axis and the coordinates ‘x = a’ and ‘x = b’. If ‘x’ is a point in [a, b], then ∫_{a}^{x} f(x) dx represents the area of the shaded region {A(x)} in the figure above.

Note: We are assuming that f(x)> 0 for x ∈ [a, b]. However, the assertions made below are true for other functions as well. The area of this shaded region depends on the value of ‘x’. Or, it is a function of ‘x’. This is denoted by A(x) and represented as follows:

A(x) = ∫_{a}^{x} f(x) dx … (1)

This helps us define the two basic fundamental theorems of calculus.

## First Fundamental Theorem of Calculus

If ‘f’ is a continuous function on the closed interval [a, b] and A (x) is the area function. Then A′(x) = f (x), for all x ∈ [a, b].

## Second fundamental theorem of Calculus

If ‘f’ is a continuous function defined on the closed interval [a, b] and F is an anti-derivative of ‘f’. Then

∫_{a}^{b} f(x) dx = [F(x)]^{a}_{b} = F(b) – F(a)

### Important Points to Remember

- In words, the Theorem 2 tells us that ∫
_{a}^{b}f(x) dx = (value of the anti-derivative ‘F’ of ‘f’ at the upper limit b) – (value of the same anti-derivative at the lower limit a). - This theorem is useful because we can calculate the definite integral without calculating the limit of a sum.
- The most important step in evaluating a definite integral is finding a function whose derivative is equal to the integrand. It strengthens the relationship between differentiation and integration.
- In ∫
_{a}^{b}f(x) dx, the function ‘f’ should be well defined and continuous in [a, b].

**You can download Integrals Cheat Sheet by clicking on the download button below**

## Calculation Steps

Two simple steps can help you calculate ∫_{a}^{b} f(x) dx as shown below:

- Find the indefinite integral ∫ f(x) dx. Let this be F(x). There is no need to keep the integration constant C because it disappears while evaluating the value of the definite integral.
- Evaluate F(b) – F(a) = [F (x )]
_{a}^{b}. This is the value of ∫_{a}^{b}f(x) dx.

That’s it! Let’s look at some examples now.

### Example 1

Evaluate ∫_{2}^{3} x^{2} dx

Solution: Let, I = ∫_{2}^{3} x^{2} dx.

Now, the indefinite integral, ∫ x^{2} dx = x^{3}/3 = F(x)

Using the second fundamental theorem of calculus, we get

I = F(a) – F(b) = (3^{3}/3) – (2^{3}/3) = 27/3 – 8/3 = 19/3

Therefore, ∫_{2}^{3} x^{2} dx = 19/3.

### Example 2

Evaluate ∫_{4}^{9} [√x / (30 – x^{3/2})^{2}] dx

Solution: Let I = ∫_{4}^{9} [√x / (30 – x^{3/2})^{2}] dx

First, we find the anti-derivative of the integrand.

Let’s substitute (30 – x^{3/2}) = t, so that – 3/2 √x dx = dt or √x dx = – 2/3 dt.

Therefore, the indefinite integral

∫ [√x / (30 – x^{3/2})^{2}] dx = – 2/3 ∫ dt/t^{2} = 2/3 (1/t)

= 2/3 [1 / (30 – x^{3/2})] = F(x).

Hence, using the second fundamental theorem of calculus, we get

I = F(9) – F(4) = 2/3 [1 / (30 – 27) – 1 / (30 – 8)] = 2/3 [1/3 – 1/22] = 19/99

Therefore,

∫_{4}^{9} [√x / (30 – x^{3/2})^{2}] dx = 19/99

### Example 3

Evaluate ∫_{1}^{2} xdx / (x + 1)(x + 2)

Solution: Let I = ∫_{1}^{2} x dx / (x + 1)(x + 2)

Using the rules of partial fractions, we get

x / (x + 1)(x + 2) = – 1 / (x + 1) + 2 / (x + 2)

Therefore, the indefinite integral

∫ x dx / (x + 1)(x + 2) = – log |x + 1| + 2log |x + 2| = F(x)

Hence, using the second fundamental theorem of calculus, we get

I = F(2) – F(1) = [– log 3 + 2log 4] – [– log 2 + 2log 3]

= – 3log 3 + log 2 + 2log 4 = log (32/27)

Therefore,

∫_{1}^{2} xdx / (x + 1)(x + 2) = log (32/27)

### Example 4

Evaluate ∫_{0}^{π/4} sin^{3}2t cos2t dt

Solution: Let I = ∫_{0}^{π/4} sin^{3}2t cos2t dt

To solve the indefinite integral, let’s substitute sin2t = u, so that 2 cos2t dt = du or cos2t dt = ½ du. Therefore,

∫ sin^{3}2t cos2t dt = ½ ∫ u^{3} du = (1/8) u^{4} = (1/8) sin^{4}2t = F(x).

Hence, using the second fundamental theorem of calculus, we get

I = F(π/4) – F(0) = 1/8 [sin^{4} (π/2) – sin^{4} 0] = 1/8(1 – 0) = 1/8

Therefore, ∫_{0}^{π/4} sin^{3}2t cos2t dt = 1/8.

## More Solved Examples for You

**Question 1:** **Evaluate ∫ _{0}^{1} dx / (√[1 – x^{2}])**

**Answer :** Let I = ∫_{0}^{1} dx / (√[1 – x^{2}]). Let’s find the indefinite integral.

∫ dx / (√[1 – x^{2}]) = ∫ dx / (√[1^{2} – x^{2}])

Using the standard integral ∫ dx / (√[a^{2} – x^{2}]) = sin^{–1} (x/a), we get

∫ dx / (√[1 – x^{2}]) = sin^{–1} (x/1)

= sin^{–1} (x/1) = sin^{–1}x = F(x)

Hence, using the second fundamental theorem of calculus, we get

I = F(1) – F(0) = sin^{–1}1 – sin^{–1}0 = π/2 – 0 = π/2

Therefore, ∫_{0}^{1} dx / (√[1 – x^{2}]) = π/2

**Question 2:** **Evaluate ∫ _{-1}^{1} (x + 1) dx**

**Answer :** Let I = ∫_{-1}^{1} (x + 1) dx. Let’s find the indefinite integral.

∫ (x + 1) dx = (x^{2}/2) + x = F(x)

Hence, using the second fundamental theorem of calculus, we get

I = F(1) – F(- 1) = [(1/2) + 1] – [(-1/2) + (- 1)]

= ½ + 1 – ½ + 1 = 2

Therefore, ∫_{-1}^{1} (x + 1) dx = 2

**Question 3:** Evaluate ∫_{0}^{1} (xe^{x} + sin (πx/4) dx

**Answer :** Let I = ∫_{0}^{1} (xe^{x} + sin (πx/4) dx. To solve the indefinite integral, let

∫ (xe^{x} + sin (πx/4) dx = ∫ x e^{x} dx + ∫ sin (πx/4) dx = I_{1} + I_{2} … (A)

Where, I_{1} = ∫ x e^{x} dx and I_{2} = ∫ sin (πx/4) dx

#### Solving I_{1}

To solve I_{1}, we will use the rule of integration by parts. According to this rule,

∫ f(x) g(x) dx = x ∫ g(x) dx – ∫ [f’(x) ∫ g(x) dx] dx

Let, The first function = f(x) = x and the second function = g(x) = e^{x}. Therefore,

I_{1} = ∫ x e^{x} dx = x ∫ e^{x} dx – ∫ [(dx/dx) ∫e^{x} dx] dx

= x e^{x} – ∫ [1.e^{x} dx] dx = xe^{x} – e^{x} = e^{x} (x – 1)

Therefore, I_{1} = e^{x} (x – 1)

#### Solving I2

I_{2} = ∫ sin (πx/4) dx = 1/( π/4) [– cos (πx/4)] = – 4/π cos (πx/4)

Substituting the values of I1 and I2 in equation (A), we get,

∫ (xe^{x} + sin (πx/4) dx = I_{1} + I_{2} = e^{x} (x – 1) – 4/π cos (πx/4) = F(x)

#### Solving I

Hence, using the second fundamental theorem of calculus, we get

I = F(1) – F(0)

= [e^{1} (1 – 1) – 4/π cos (π/4)] – [e^{0} (0 – 1) – 4/π cos 0]

= [– 4/π cos (π/4)] – [(– 1) – 4/π] = –4/π cos (π/4) + 1 + 4/π

= –4/π (1/√2) + 1 + 4/π = 1 + 4/π – 2√2/ π

Therefore, ∫_{0}^{1} (xe^{x} + sin (πx/4) dx = 1 + 4/π – 2√2/ π

**Question 4: State the fundamental theorem of calculus part 1?**

**Answer:** The fundamental theorem of calculus part 1 states that the derivative of the integral of a function gives the integrand; that is distinction and integration are inverse operations. In addition, they cancel each other out. Moreover, the integral function is an anti-derivative.

**Question 5: State the fundamental theorem of calculus part 2?**

**Answer:** As per the fundamental theorem of calculus part 2 states that it holds for ∫a continuous function on an open interval Ι and a any point in I. Furthermore, it states that if F is defined by the integral (anti-derivative).

**Question 6: Are anti-derivatives and integrals the same?**

**Answer:** Definite integral gives you the integral among a and I and some point. On the other hand, indefinite integral gives you the integral between a and I at some indefinite point that represented by the variable x. However, the fundamental theorem of calculus says that anti-derivatives and indefinite integrals are the same things.

**Question 7: Why is the anti-derivative the area under the curve?**

**Answer: **If the integration of a function f(x), you get anti-derivative F(x). In addition, if you evaluate the anti-derivative over a specific domain [a, b], then you get the area under the curve.

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