Integral

To solve the problem, we need to evaluate the integral , set it equal to , and find an integer root of the equation .

1. Evaluate the Integral:
Compute .
Antiderivative: .
Evaluate at the limits:
.
Thus, .

2. Form the Equation:
Substitute into :
.
So, , which simplifies to:
, or .

3. Solve for the Integer Root:
Solve using factoring:
Find numbers whose product is and sum is . These are and .
Rewrite:
.
Group:
, so .
Roots: , or .
The integer root is .

Final Answer:
The integer root is .

To solve the problem, we need to find the value of for the normal line at the point on the parametric curve defined by and .

1. Find the Parameter at :
At , set and .
Solve :
, so or .
Solve :
, so or .
The common solution is .

2. Compute the Derivative :
Find and .
At :
, .
Thus, .

3. Determine the Slope of the Normal:
The slope of the tangent is , so the slope of the normal is .

4. Find the Equation of the Normal:
At , the normal’s equation is .
Multiply through by 8:
.
Rearrange:
.

5. Evaluate :
From , we have , , .
The GCD of is 1, confirming the equation is in standard form.
Compute .

Final Answer:
The value of is .

To solve the problem, we need to find the maximum value of the function and the corresponding .

1. Find the Critical Points:
Compute the first derivative:
.
Set :
. Since , we have , so .

2. Determine the Nature of the Critical Point:
Compute the second derivative:
.
Evaluate at :
, indicating a maximum.

3. Calculate the Maximum Value:
At , the function value is:
.
Thus, and , so .

Final Answer:
The maximum occurs at .

To solve the problem, we need to evaluate the integral .

1. Split the Integral:
Since for and for , we have:
for and for .
Thus:
.

2. Evaluate the Integrals:
Compute each part:
.
.
Sum:
.

3. Alternative Approach (Symmetry):
Define . Check if it’s odd:
, so is odd.
For an odd function, .
Thus, .

Final Answer:
The value of the integral is .

To solve the problem, we need to evaluate the integral

1. Split the Integral:
Write .

2. Analyze the First Integral:
Let .
Check if is odd:
.
Since is odd, .

3. Analyze the Second Integral:
Let .
Check if is even:
.
Since is even, .

4. Evaluate the Remaining Integral:
Thus, .
Use :
.

5. Substitute and Integrate:
Let , so . Limits: , .
Then .
Integrate:
.

Final Answer:
The value of the integral is .

To solve the problem, we need to find the area under the curve over the interval .

1. Set Up the Integral:
Since is non-negative, the area is:
.

2. Use Periodicity:
The period of is , so has period .
Thus, the integral over (four periods) is:
.
Since on , there.

3. Evaluate the Integral:
Compute:
.
Simplify:
.

Final Answer:
The area is .

Step 1: Understanding the Given Integral We are given: Using the standard integral formula: where is the integral of , we identify:

Step 2: Compute Integrating each term: Thus,

Step 3: Evaluate Taking LCM (12): Thus, .

Step 1: Simplify the Denominator We are given: Expanding the denominator: Using the identity : Rewriting: Approximating for large powers: Using the identity: Thus, the integral simplifies to:

Step 2: Evaluating the Integral Taking LCM (10): Since the given form requires division by 2, the final answer is:

Step 1: Substituting Trigonometric Identities We rewrite using the sine and cosine functions: Multiplying numerator and denominator by , we obtain: Thus, the given integral becomes:

Step 2: Using Substitution Let: Differentiating both sides: We rewrite the integral using substitution: Using standard integral properties:

Step 1: Using a Suitable Substitution Let: Differentiating both sides: Rewriting the denominator: Thus, the integral transforms into:

Step 2: Evaluating the Integral The fraction simplifies to: Using the inverse hyperbolic sine transformation: