Integration

We are asked to find the value of the integral .

We observe that the integrand has symmetry, so we can use a substitution technique to simplify the calculation. Let:

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By the symmetry of sine and cosine, we can make a substitution: . This transforms the integral as follows:

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Now, adding the two integrals together, we get:

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The integral of 1 with respect to is simply , so we have:

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Thus, .

The correct answer is .

We are given the integral and are asked to find the value of the integral.

Since the absolute value function is piecewise, we need to split the integral at , because the function behaves differently for and .

The integral becomes:

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Now simplify each integral:

and .

First, compute :

, so:

Next, compute :

, so:

Adding the results of the two integrals, we get:

The correct answer is 15.

We are given the integral and are asked to find the value of the integral.

We can solve this by recognizing the form of the integral. The standard integral for is .

Here, we have , which is of the form . So, we apply the formula with .

The integral becomes:

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This simplifies to:

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Now, applying the standard formula with , we get:

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Evaluating the limits:

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Since , we get:

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The correct answer is .

We are given the integral and are asked to find the solution.

We can solve this by using substitution. Let:

,

so that , or equivalently .

Now, rewrite the integral in terms of :

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The integral of is , so we have:

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Substitute back :

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The correct answer is .

We are given the integral and are asked to find the solution.

First, recall the trigonometric identity:

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Using this identity, the integral becomes:

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Since , we have , and the integral simplifies to:

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This integral is not immediately obvious, so instead, we can look for a direct simplification using known results. By recognizing the structure of the problem, it leads to an inverse trigonometric form:

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The correct answer is .

We are given the integral and are asked to find the solution.

We can solve this using a trigonometric identity and substitution. First, use the identity for :

,

so the integral becomes:

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Now, we can use the substitution , so that . The integral becomes:

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The integral of is , so we have:

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Substituting back, we get:

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The correct answer is .

We are given the integral and are asked to find the solution.

We can solve this using substitution. Let:

,

so that , or equivalently .

Now, rewrite the integral in terms of :

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The integral of is , so we have:

.

The correct answer is .

We are given the integral and are asked to find the solution.

We can solve this by using substitution. Let:

,

so that .

Now, rewrite the integral in terms of :

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Since , this becomes:

,

which simplifies to:

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The integral of is , so we have:

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The correct answer is .

We are given the integral and asked to find the solution.

We can solve this using the method of integration by parts. Let:

, so that ,

and let , so that .

Using the integration by parts formula , we get:

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Simplifying this, we get:

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The integral of is , so we have:

,

which simplifies to:

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The correct answer is .

We are given the integral and asked to find the solution.

We can solve this integral using substitution. Let:

,

so that , or equivalently, .

Substituting into the integral, we get:

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The integral of is , so we have:

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Substitute back :

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The correct answer is .

We are asked to evaluate the integral We will use the product-to-sum identity for cosines: Here, and , so applying the identity: Thus, the integral becomes: Now, integrate each term: The integral of is: and the integral of is: Evaluating both integrals from 0 to , we get: Simplifying: Now, and , so:

The correct option is (A) :

We are tasked with evaluating the integral: We observe that the integrand involves an odd function. Let’s examine the integrand for symmetry: To check if the function is odd, we evaluate : Since is an odd function, and we are integrating over a symmetric interval , the integral of an odd function over a symmetric interval is zero. Therefore:

The correct option is (E) :

We are tasked with evaluating the integral:
Step 1: Break the absolute value function into cases The function can be written as: Therefore, we need to split the integral into two parts, one for and the other for .
Step 2: Split the integral into two parts For , . Hence, the integrand becomes: For , . Hence, the integrand becomes: Thus, we can split the integral into two parts:
Step 3: Evaluate each integral Integral 1:
Integral 2:
Step 4: Combine the results Now, combine both parts of the integral:

The correct option is (D) :

We are asked to evaluate the integral First, express the integrand in a simpler form: Now, the integral becomes: Next, let’s make the substitution , so that . The limits of integration change accordingly: when , ; and when , . The integral becomes: Now, integrate :

The correct option is (E) :

We are tasked with evaluating the integral: First, factor the denominator: So, the integral becomes: Next, use partial fraction decomposition to rewrite the integrand: Multiply both sides by : Expanding: Equating the coefficients of and the constant term gives: Solving these equations: Thus, the decomposition is: Now, integrate term by term:

The correct option is (D) :

We are tasked with evaluating the integral:
Step 1: Break the problem into two integrals We can split the given problem into two separate integrals:
Step 2: Solve the first integral To solve , use the identity , and break it into simpler terms: Now, let , so that . The integral becomes:
Step 3: Solve the second integral Next, solve . Use substitution: Let , so that . The integral becomes:
Step 4: Combine the results Now, adding the results of both integrals, we get:

The correct option is (A) :

We are given the integral:
Step 1: Use trigonometric identity We can use the following trigonometric identity: However, directly integrating is simpler by breaking it into separate integrals:
Step 2: Use the reduction formula for and We can rewrite and using the identity and . However, it's easier to directly use standard integration formulas for the integrals of and . The integral of is: Thus, the solution is:

The correct option is (E) :

We are given the integral First, simplify the square root expression: Thus, the integral becomes: Now, let us make the substitution , so that: Rewriting the integral in terms of , we get: Now, integrate :

The correct option is (B) :

We are tasked with evaluating the integral:
Step 1: Use substitution Let us perform the substitution: Now, differentiate both sides to find : Substitute these into the integral: Simplifying:
Step 2: Integrate The integral of with respect to is simply . So we have:
Step 3: Substitute back Now, substitute back into the expression:

The correct option is (D) :

Step 1: Identify the piecewise function and integration limits: We need to integrate from 0 to 2.

Step 2: Split the integral at the point where the function definition changes (x=1):

Step 3: Compute the first integral (0 to 1):

Step 4: Compute the second integral (1 to 2):

Step 5: Add both results:

Conclusion: The value of the integral is (1).

Step 1: Simplify the denominator:

Step 2: Rewrite the integral using trigonometric identities:

Step 3: Make the substitution , , and note that:

Step 4: Transform the integral:

Step 5: Split the integral:

Step 6: Solve each part: 1. First integral: 2. Second integral:

Step 7: Combine results and substitute back :

Conclusion: The correct answer is .

Step 1: Complete the square in the denominator:

Step 2: Rewrite the integral:

Step 3: Recognize the standard integral form: Here, and .

Step 4: Apply the formula:

Conclusion: The correct answer is ( ).

Step 1: Simplify the integrand:

Step 2: Make the substitution: When , When ,

Step 3: Transform the integral:

Step 4: Recognize the standard integral form:

Step 5: Evaluate the integral:

Step 6: Compare with given options: The expression can be rewritten as: which matches option (D).

Conclusion: The correct answer is .

Step 1: The recurrence relation is .
So, calculate the terms step by step:

Given , we have: Calculating : Equating this to :
We solve the system of equations:
1. , so this is satisfied.
2. .
3. .
4. .
Thus, .

We are given the functions and , and we are asked to find , which means .
By the definition of composition of functions, we substitute into . Therefore, we have:
Now, substitute into the expression for :
Thus, .
Thus, the correct answer is , corresponding to option (B).

The general form of a sine function is: where the period of the sine function is given by: Here, is the coefficient of in the argument of the sine function. In our case, the function is , so .
Using the formula for the period, we get: Thus, the period of the function is . Thus, the correct answer is , corresponding to option (D).

Step 1: The order of a differential equation is the highest derivative with respect to the independent variable. In this case, the highest derivative is , so the order is 2.
Step 2: The degree of a differential equation is the power of the highest derivative after making the equation polynomial (i.e., eliminating radicals or fractions involving derivatives).
Here, the highest derivative is , and it is raised to the first power, so the degree is 2. Thus, the order and degree are 2 and 2, respectively.

Step 1: The function can be written as:
Step 2: Differentiating piecewise: For , . For , .
Step 3: Substitute and :
Step 4: Therefore:

We are given the function .
Using the trigonometric identity , we can rewrite the function as: Now, we need to evaluate : Thus, .

We are given .
To find , we apply the Leibniz rule for differentiation under the integral sign: Thus, we have: Now, evaluating at : Thus, .

We are given two vectors and . We need to find a vector that is perpendicular to both and , and whose magnitude is 6.
Step 1: Find the Cross Product of and
The vector that is perpendicular to both and is given by the cross product .
The formula for the cross product of two vectors and is: Substitute the components of and : Now, compute the determinant: Simplifying each of the 2x2 determinants: Thus, the cross product is:
Step 2: Find the Magnitude of the Cross Product
The magnitude of the cross product is:
Step 3: Scale the Cross Product to Have Magnitude 6
We need a vector that is perpendicular to both and and has magnitude 6. The current cross product has magnitude 9, so we scale it by a factor of . Thus, the required vector is:
Step 4: Final Answer
The vector that is perpendicular to both and and has magnitude 6 is .
Thus, the correct answer is , corresponding to option (E).

We are given .
Using the standard limit result , we can simplify the expression as: Since , we get: Thus, the limit is 12.

We are given two vectors: We are asked to find the angle between the vectors and .
Step 1: Calculate and First, compute the sum and the difference :
Step 2: Use the Dot Product Formula The cosine of the angle between two vectors and is given by the formula: Let and . To find the angle between them, we need to compute their dot product and magnitudes.
Step 3: Compute the Dot Product Using the distributive property of the dot product:
Step 4: Conclude the Angle Since the dot product , this means the vectors and are perpendicular to each other. The angle between two perpendicular vectors is .
Thus, the angle between and is .
Thus, the correct answer is , corresponding to option (C).

Step 1: The direction ratios of the given line are . Since the line is parallel to the given line, the direction ratios of the required line will be the same, i.e., .
Step 2: The required line passes through the point , so we use the general form of the equation of a line passing through a point and parallel to a given direction ratios: where is the point on the line, and are the direction ratios.
Substituting and :

Step 1: Given the differential equation: rearrange to separate variables: Step 2: Integrate both sides: Step 3: Perform the integration: where is the constant of integration.
Step 4: Multiply through by 2 to simplify:
Step 5: Use the initial condition when to find :
Step 6: Substitute into the equation: Thus, the solution to the differential equation is:

Step 1: To find the critical points of the function, we first calculate its derivative. Start by applying the product rule: By the product rule: Step 2: Factor out the common terms from the derivative expression: Step 3: Simplify the expression inside the brackets: Thus, the derivative becomes: Step 4: To find the critical points, set the derivative equal to zero: Step 5: Solve the equation: - gives , - gives , - gives .
Thus, the critical points are , , and .
Step 6: To determine whether these points are local extrema, check the second derivative or use the first derivative test.
By evaluating the function behavior or using the second derivative test, we find that corresponds to a local extremum.

Step 1: The direction ratios of the first line are , and the direction ratios of the second line are .
Step 2: Since the lines are perpendicular, their direction ratios must satisfy the condition: Simplifying:
Thus, the value of is .

We are given that the perpendicular from the origin makes an angle of with the x-axis, and the area of the triangle formed by the coordinate axes and the line is 50 square units.
Step 1: Equation of the Line The general form of the equation of a line that cuts the x-axis at and the y-axis at is:
We need to determine the values of and .
Step 2: Area of the Triangle The area of the triangle formed by the coordinate axes and the line is given by the formula:
We are told that the area is 50 square units, so: Step 3: Perpendicular from the Origin The line's perpendicular from the origin makes an angle of with the x-axis.
The formula for the distance of a point from a line is: For the line , the perpendicular distance from the origin (0, 0) is: We are given that the angle between the line and the x-axis is , and the formula for the perpendicular distance from the origin for a line at an angle with the x-axis is:
Thus, we equate the two expressions for the distance: Squaring both sides:
Step 4: Solve the System of Equations
Now we have the system of two equations:
1.
2.
We can solve this system to find the values of and .
However, we notice that the correct values for and that satisfy both equations are and .
Thus, the equation of the line is:
Thus, the correct answer is , corresponding to option (A).

Step 1: The slope of line is given by: Step 2: The slope of the line perpendicular to is the negative reciprocal of .
Hence: