(9)

Step 1:

The integral is .

Continuity Implies Integrability Theorem:

If a function is continuous on the closed interval , then is integrable on .

exists.

Consider .

The function is is not continuous on the interval .

is exists.

Step 2:

Evaluation of integral:

The integral is

Apply sum and difference rule in intgration: .

Apply power rule in integration: .

.

.

Solution:

.

(10)

The integral is . \ \

Continuity Implies Integrability Theorem:

If a function is continuous on the closed interval , then is integrable on .

exists.

Consider the function . \ \

The function is continuous on the interval .

is exists.

Step 2:

Evaluation of integral:

The integral is .

Apply power rule in integration: .

.

.

Solution:

.

(11)

Step 1:

The integral is . \ \

Continuity Implies Integrability Theorem:

If a function is continuous on the closed interval , then is integrable on .

exists.

Consider the function . \ \

The function is continuous on the interval .

is exists.

Step 2:

Evaluation of integral:

The integral is .

Apply special integration formula: .

In this case and .

.

.

Solution:

.

.