Unit 5: Integration II - Subjective Questions

Integration by parts is used to integrate the product of two functions.

Derivation:

If and are two differentiable functions of , then by the product rule of differentiation:

Integrating both sides with respect to :

Rearranging gives the formula for integration by parts:

Or equivalently, if is the first function and is the second function:

ILATE Rule for choosing the first function (the one to be differentiated), in order of preference:

• I – Inverse trigonometric functions
• L – Logarithmic functions
• A – Algebraic functions
• T – Trigonometric functions
• E – Exponential functions

The function appearing earlier in ILATE is taken as the first function and the other as the second function.

We use the formula:

Choosing functions (ILATE): Algebraic () is first, exponential () is second.

• Let
• Let

Applying the formula:

Result:

Choosing functions (ILATE): Logarithmic () is first, algebraic () is second.

• Let
• Let

Applying :

Result:

Definition:

If , then the definite integral of from to is defined as:

where is called the lower limit and is the upper limit of integration. This is known as the Fundamental Theorem of Calculus.

Geometrical Interpretation:

• The definite integral represents the area bounded by the curve , the -axis, and the ordinates and .
• If on , the area lies above the -axis and is positive.
• If on , the area lies below the -axis and the integral gives a negative value.
• Unlike the indefinite integral, the definite integral gives a definite numerical value (no constant of integration ).

Definite integral as the limit of a sum:

Let be a continuous function defined on the closed interval . Divide the interval into equal subintervals, each of width:

The points of division are:

Consider the sum of areas of rectangles:

As , the width , and the sum approaches the exact area under the curve. Therefore:

where .

This is the definition of the definite integral as the limit of a sum.

Here , , , so .

Using:

Here , so:

Therefore:

Using and :

As , :

Result:

The main properties of definite integrals are:

1. (integral independent of variable of integration)

2. (interchanging limits changes sign)

3. , where

4.

5.

6. (equal limits give zero)

7.

8.

To prove:

Proof:

Let .

Substitute , so that .

Changing the limits:

• When ,
• When ,

Therefore:

Using the property :

Since the variable of integration is a dummy variable, replace by :

Hence proved.

Let

Using the property with :

Since and :

Adding (1) and (2):

Result:

We use the standard result:

Identifying the function:

Let , then .

The given integral becomes:

Applying the result directly:

Result:

Choosing functions (ILATE): Algebraic () is first, trigonometric () is second.

• Let
• Let

Applying :

Now evaluate (integration by parts again):

• Let
• Let

Substituting back into (1):

Result:

We first find the antiderivative:

Applying the limits from to :

At upper limit :

At lower limit :

Therefore:

Result:

The differences between indefinite and definite integrals are:

Indefinite Integral:

• Written as with no limits.
• The result is a function of plus an arbitrary constant .
• Represents a family of curves (antiderivatives).
• General form:
• Does not give a numerical value.

Definite Integral:

• Written as with lower limit and upper limit .
• The result is a definite numerical value (a constant).
• No arbitrary constant appears in the answer.
• Represents the area under the curve between and .
• Evaluated as .

Key relation: The definite integral is evaluated by first finding the indefinite integral (antiderivative) and then substituting the limits.

Proof:

Using the property:

Consider the first integral .

Substitute , so .

• When , ; when , .

Substituting into (1):

Case 1: is even, so :

Case 2: is odd, so :

Hence proved.

We write , treating as the second function.

Choosing functions:

• Let
• Let

Applying :

Result:

Let

Using the property with :

Adding (1) and (2):

Result:

Choosing functions (ILATE): Algebraic () is first, trigonometric () is second.

• Let
• Let

Applying :

Steps in integration by parts:

1. Identify the two functions in the product.
2. Choose the first function () and second function () using the ILATE rule.
3. Differentiate the first function to get .
4. Integrate the second function to get .
5. Substitute into the formula .
6. Evaluate the remaining integral and add the constant .

Result:

First find the indefinite integral using integration by parts.

• Let
• Let

Now apply the limits from to :

At :

At :

Therefore:

Result:

For evaluating as the limit of a sum, we use the alternative approach. Here , , .

Alternatively, this integral is straightforward using the fundamental theorem, and the limit of a sum gives the same result. Using the antiderivative approach to confirm:

At upper limit :

At lower limit :

Therefore:

Note: The limit-of-a-sum definition with yields the same value , confirming consistency between the two methods.

Result:

The Fundamental Theorem of Calculus establishes the relationship between differentiation and integration, showing they are inverse operations. It has two parts:

First Fundamental Theorem (Integral Function):

If is a continuous function on , and we define the area function:

then is differentiable and:

This means differentiating the integral of a function returns the original function.

Second Fundamental Theorem (Evaluation):

If is continuous on and is an antiderivative of (i.e., ), then:

Connection between differentiation and integration:

• Integration is the reverse process of differentiation.
• The theorem allows us to compute definite integrals without using the limit of a sum, simply by finding an antiderivative and evaluating it at the limits.
• It bridges the concept of area under a curve (integration) with the concept of rate of change / slope (differentiation).

This theorem is the foundation that makes practical evaluation of definite integrals possible.