Unit 3: Fundamentals of integral calculus - Subjective Questions

Linearity rules:

• Constant multiple rule:
• Sum and difference rule:

Power rule:
For any real number ,

The restriction is necessary because division by would otherwise be undefined. For ,

Here, is the constant of integration.

Integrate each term separately using linearity and the power rule:


Therefore,

After simplification,

Differentiating this result reproduces the original integrand.

Use the standard formulas

and

Applying linearity,


Hence,

The answer can be checked by differentiation.

An indefinite integral represents a family of antiderivatives. If , then

for every constant , because the derivative of a constant is zero.

For example,

The functions , , and all have derivative . Thus, omitting would give only one member of the complete family of antiderivatives.

In a definite integral, constants cancel when the limits are applied:

Integration by substitution reverses the chain rule. If

then

Procedure:

• Select an inner expression and set it equal to .
• Compute .
• Rewrite the entire integral in terms of .
• Integrate with respect to .
• Substitute the original expression back for .

The method follows from

Thus, substitution is particularly useful when the integrand contains a composite function together with its derivative, possibly differing by a constant factor.

Let

Then

The integral becomes

Substituting gives

Since for every real , this may also be written as

Indeed,

Choose

so that

Therefore,

Using the power rule,

Substituting back,

Let

Thus, . Change the limits:

• When , .
• When , .

Hence,

Now,


Because the limits were converted to -values, no back-substitution was required.

Start with the product rule:

Multiplying by gives

Integrating both sides,

Since , rearrangement gives

For definite integrals, the corresponding formula is

A useful guideline for choosing is LIATE: logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential functions, in that order of preference.

Use

Choose

Then

Therefore,


Thus,

Verification:

Choose the logarithmic function as :

Then

Applying integration by parts,

Thus,


Hence,

Write

Choose

Then

Integration by parts gives

Use :

Therefore,

Hence, the reduction formula is

Partial fractions are used to integrate a rational function

when . If the fraction is improper, polynomial division must be performed first.

Forms of decomposition:

• Distinct linear factors:
• A repeated linear factor :
• An irreducible quadratic factor :
• If the quadratic factor is repeated, one such linear numerator is included for every power of the factor.

After determining the constants, each simpler fraction is integrated using logarithmic, power, or inverse-trigonometric formulas.

Assume

Multiplying by gives

Set :

Set :

Thus,

Integrating,

Because is repeated, write

Multiplying by gives

Setting gives , and setting gives . Comparing the coefficient of gives , so .

Therefore,

Integrating term by term,

The derivative of the denominator is . Rewrite the numerator as

Thus,

The first integral is

Complete the square in the second:

Therefore,

Hence,

Important properties of definite integrals include:

• Identical limits:
• Reversal of limits:
• Additivity over intervals:
• Linearity:
• Comparison: If on , then

These properties follow from interpreting a definite integral as signed area or from the Fundamental Theorem of Calculus.

Let

Use the substitution

When , , and when , . Therefore,

Renaming as gives

Adding the two equal representations of ,

Hence,

This identity often simplifies integrals over symmetric intervals.

For an even function, , and

For an odd function, , and

In the given integrand, is odd, while is even. Therefore,

Thus,

Let

Using ,

Adding the two expressions,

Since ,

Put . Then

By symmetry about ,

Hence the transformed integral equals , so

Therefore,