Integration

The equation of the ellipse is:
Rearrange to solve for :

Step 1: Use symmetry to simplify the calculation. The ellipse is symmetric about the -axis. The area between and can be calculated as twice the area above the -axis:

Step 2: Change the limits and integrate. Since the integrand is even (symmetric about the -axis), we can further simplify:

Step 3: Substitution for simplification.
Let: When , , and when , .

The integral becomes: Simplify:

Step 4: Evaluate the integral. The integral of is: Evaluate from to :
Simplify:
Thus:

Step 5: Final area. Substitute back into the expression for the area:

Final Answer:

Step 1: Understand the problem
The curve represents the upper semicircle with radius and center at the origin. We are tasked to find the area of the region enclosed between this curve, the lines , , and the -axis.
Step 2: Set up the integral
The required area can be calculated as twice the area between the curve and the -axis for (using symmetry): Step 3: Evaluate the integral using substitution
Let , so , and for , , and for , , giving . Substituting: Simplify: Use the double-angle identity : Split into two integrals: Evaluate each term: 1. For : 2. For : Using : Step 4: Combine results
The total area is: Conclusion:
The area of the region is:

Step 1: Simplify the integrand
The given integral is: Use the identity . Substituting: Step 2: Integrate
The integral of is: Substituting this, we have: Step 3: Evaluate the limits
At : At : Substitute these values: Step 4: Simplify the result

Step 1: Simplify the integrand
Use the identity and : Step 2: Rewrite the integral
Step 3: Integrate term by term
For , use substitution : For , combine it with the first term:

Step 1: Substitute
Let , so and . Substituting into the integral: Step 2: Simplify the integrand
Rewrite using partial fraction decomposition. We have: Step 3: Integrate each term
The integral becomes: 1. For , the result is: 2. For , rewrite using partial fractions: Integrating term by term: Step 4: Combine results and back-substitute
Substitute back :
Conclusion: The solution to the integral is:

Step 1: Find the derivative of
The given function is . Differentiate: Step 2: Find critical points
Set : Step 3: Evaluate at critical points and endpoints
At : At :
Step 4: Conclude the result
The absolute maximum value is at , and the absolute minimum value is at .

Step 1: The given integral can be written as: Let:

Step 2: Now, calculate the derivative of : Simplify the numerator: Thus, the integral becomes:

Step 3: Using the standard result: Substitute :

Final Answer:

Explanation: 1. Splitting the Integral: The given integral is split into terms containing and .

2. Defining : The function is chosen as because its derivative results in the second term, .

3. Applying the Formula: The integral formula for is directly applied.

4. Substitution: Finally, substituting into the formula gives the result.

Step 1: Splitting the integral. The absolute value function is defined as Splitting the integral at : Step 2: Evaluating the integrals. Step 3: Summing the results. Conclusion: Thus, the value of the integral is , which corresponds to option .

We are asked to evaluate the integral: Step 1: Use the identity for
We can express as: Thus, the integral becomes: Step 2: Substitute and simplify the integral
Now, let's use the substitution method: Let , then , so we can rewrite the integral: Step 3: Finish the integration
After simplifying and solving: Thus, the solution to the integral is: Step 4: Correct Answer:
The correct solution to the integral is:

Step 1: Recall the direction cosine property
The sum of the squares of the cosines of the angles a line makes with the coordinate axes is: This is a fundamental property of direction cosines.
Step 2: Check each option
Option (A): True, as it is the direction cosine property.
Option (B): True, as .
Option (C): True, derived from trigonometric identities for direction cosines.
Option (D): False, as in general.
Step 3: Conclude the result
Option (D) is not true.

Step 1: Use the property of definite integrals
The integral if the function is odd. A function is odd if:
Step 2: Simplify the integral for odd functions
For odd functions:
Since , the integral over cancels with the integral over , resulting in:
Step 3: {Conclude the result}
The given integral equals 0 when .

Step 1: The given integral can be written as: Let: Step 2: Now, calculate the derivative of : Simplify the numerator: Thus, the integral becomes: Step 3: Using the standard result: Substitute : Final Answer: Explanation: 1. Splitting the Integral: The given integral is split into terms containing and .
2. Defining : The function is chosen as because its derivative results in the second term, .
3. Applying the Formula: The integral formula for is directly applied.
4. Substitution: Finally, substituting into the formula gives the result.

The curve is , which is the upper half of a parabola. The area under this curve from to can be found using the following definite integral: We know that: So, the integral becomes: Now, we can integrate: Substituting the limits of integration: Since , we get: Therefore, the area of the region is

We are asked to find the area of the region bounded by the curve , the lines , , and the x-axis ( ).

Step 1: Understanding the curve The function is a V-shaped curve. The absolute value function creates a piecewise function:
Step 2: Set up the integral We need to calculate the area between the curve and the x-axis from to . We break this integral into two parts based on the piecewise definition of the function: 1. From to , the equation for the curve is . 2. From to , the equation for the curve is . Thus, the area is given by:
Step 3: Calculate the first integral Simplifying:
Step 4: Calculate the second integral Simplifying:
Step 5: Total area Adding the two areas together:
Thus, the total area of the region is .

To evaluate:

Step 1: Use integration by parts

Let:

Using the formula:

Apply the limits:

Evaluate the first term:

Evaluate the second term:

Final Answer:

Area Bounded by the Curves

Given:

Find the area of the region bounded by the curves:

Step 1: Find points of intersection

Equate and : So the region of intersection lies between and , where:

Step 2: Set up the integral

Area between curves from to is: Here, from to :

Step 3: Evaluate the integral

✅ Final Answer:

We use partial fraction decomposition:

Multiply both sides by :

Expand both sides:

Now, equate coefficients:

Solving:

From

Substitute in

Substitute into third equation:

Now rewrite the integral:

Final Integration:

Final Answer:

We need to handle the absolute value expressions carefully. Let’s analyze where each expression inside the absolute value changes sign:

• : changes at
• : changes at
• : changes at

So, we split the interval at :

1. Interval :

In this interval:

So the integrand becomes:

2. Interval :

In this interval:

So the integrand becomes:

Final Answer:

Step 1: Trigonometric Identity Transformation

We know the identity:

Using this in the integrand:

Also, recall that:

So:

In the interval , both and are positive, so the absolute value is not needed:

Step 2: Integrating the Expression

Now the integral becomes:

Split the integral:

Compute each part:

So:

Evaluate:

Step 3: Conclusion

The value of the integral is: