Methods Of Integration

Given the function defined as:

The function is defined on the interval . 

Analysis of the increasing and decreasing intervals: - is decreasing in - Increasing in - Decreasing in - Increasing in

The graph of is as follows:

 

The function lies in the range , and the area under the curve is given as 4.

Solving for :

Therefore, the maximum value of is .

Final Answer:

The correct value of is .

Analysis of the Piecewise Function

Given: The piecewise function is defined as:

The function is defined over the interval .

Solution:

We need to determine the intervals where is increasing or decreasing.

We can calculate the derivative of for each piece and find the intervals where the derivative is positive (increasing) or negative (decreasing).

By examining the given intervals and the derivatives, we can summarize the behavior of :

• is decreasing in
• is increasing in
• is decreasing in
• is increasing in

Conclusion:

The correct answer is 8.

First-Order Linear Differential Equation Solution

To find the value of , we start by differentiating both sides of the given equation with respect to :

Given,

Differentiating both sides using the Fundamental Theorem of Calculus and the Leibniz rule:

This implies:

Rearranging terms, we get:

Adding and dividing by , we simplify:

Let , then:

This is a first-order linear differential equation that we solve using the integrating factor method:

Identifying , the integrating factor .

Multiplying the differential equation by the integrating factor:

Which simplifies to:

The left side simplifies to , giving:

Integrating both sides:

Solving for :

Using the initial condition :

Solving for :

Thus, the function is:

Finally, we find :

Evaluating this expression, we find that the closest answer is:

Final Answer:

The correct answer is .

The solutions combine to yield the sum of solutions, which is calculated as √3 + (3√3 - 6).

The correct option is (C): 4√3-6

To solve this problem, we need to analyze the behavior of the product as .

Step 1: Analyze
For if , as , must increase because the intervals get smaller and closer to 0. Thus, can assume increasingly larger values since tends to increase as increases.

Step 2: Analyze
We know from the condition for .

Step 3: Estimation of the lower bound of
To find the behavior of as , we evaluate this integral's tendency.
Change of variables: Let , so , and consider that as . On estimation, this integral behaves like: for small .

Step 4: Evaluate
As , we substitute: leads to the approximation . As increases such that reaches , the term cancels out the , leading to: .
Ultimately, .

The accurate choices are (A), (B), and (C). Given that f′′(x)>0 and f(x)−x=0, we need to determine the number of solutions. Let g(x)=f(x)−x, which implies g′(x)=f′(x)−1 and g′′(x)=f′′(x)>0. This indicates the presence of concave possibilities

The correct options are (A),(B) and (C).