Integration

To solve the integral and find the value of , we follow these steps: Given Integral: Step 1: Simplify the Integral Let’s make a substitution to simplify the integral. Let . Then, and . Substituting into the integral: Simplify the integrand: Step 2: Perform Polynomial Division Divide by : So, Step 3: Integrate Term by Term Integrate each term separately: Step 4: Substitute Back Step 5: Compare with Given Form The given form is: Comparing coefficients: Step 6: Calculate Final Answer:

Step 1: Find a reduction formula for I_n.
I_n = =
I_n =
I_n =
I_n =
Let u = tan x, then du = sec² x dx.
When x = 0, u = 0. When x = π/4, u = 1.
I_n =
I_n =
I_n =

Step 2: Use the reduction formula to find I₁₃ + I₁₁.
Using the reduction formula, we have:
I₁₃ = =
I₁₃ + I₁₁ =
Therefore, I₁₃ + I₁₁ = .

We are given the integral

Step 1: Simplify the denominator using a standard trigonometric identity:

Step 2: The integral can be simplified further, but we can use known results for trigonometric integrals to directly obtain: Thus, the correct answer is .

We are given: We need to find the expression for:

Step 1: Differentiating the Integral
Let Differentiating both sides, Now let Thus,

Step 2: Integration
Using the identity, Thus,

Step 3: Identifying , , and
From the given format, - - -

Step 4: Computing
Since , and , Since , this simplifies to:

Final Answer: (B)

We are given two curves:


We need to find the area of the smaller region above the x-axis bounded by these curves.

Step 1: Rearranging the Equations
Circle Equation:

Rearranging,

Completing the square for the -terms,

This is a circle with center and radius .

Parabola Equation:
The given parabola is:


Step 2: Intersection Points
Equating the circle and parabola:

Since ,

Expanding the square:



Thus, or .

Step 3: Area Between the Curves
The area bounded between two curves is given by:

From the given equations:
- Upper curve (circle) →
- Lower curve (parabola) →


Step 4: Evaluating the Integrals
1. Integral for the circle:

2. Integral for the parabola:


Step 5: Computing the Final Area



Step 6: Final Answer


Final Answer: (B)

Let , so . Then,

Substituting into the integral, we get:

Therefore, the correct answer is:

Step 1: Rewrite the integral.
Let I = .
We can rewrite the integral as: I =

Step 2: Use substitution.
Let u = x^2. Then , so du = 2x dx.
We can rewrite the integral as: I =

Step 3: Recognize the derivative of a quotient.
We can rewrite the integral as: I =
Notice that this looks like the derivative of a quotient.
Let f(x) = .
Then f'(x) = .

Step 4: Integrate.
Since f'(x) = , we have:
I =
Therefore, the integral is .

Step 1: Apply the Mean Value Theorem.
Since f(x) is differentiable on [0, 2024], by the Mean Value Theorem, there exists a c (0, 2024) such that:
f'(c) =
f'(c) =
f'(c) =

Step 2: Use the given condition f'(x) 5.
Since f'(x) 5 for all x [0, 2024], we have f'(c) 5.


Step 3: Solve for f(2024).
f(2024) + 2 5 2024
f(2024) + 2 10120
f(2024) 10120 - 2
f(2024) 10118

Step 4: Determine the least possible value of f(2024).
The least possible value of f(2024) is 10118.
Therefore, the least possible value of f(2024) is 10,118.

Step 1: Use substitution to simplify the integral.
Let u = x^2. Then du = 2x dx, so x dx = du.
f(x) =

Step 2: Use partial fractions.


When u = -1, .
When u = -3, .


Step 3: Integrate with respect to u.
f(x) =
f(x) =
f(x) =
f(x) =

Step 4: Substitute back u = x^2.
f(x) =

Step 5: Use the given condition f(3) = .
f(3) =

C = 0

Step 6: Find f(0).
f(x) =
f(0) =
Therefore, f(0) = .

Step 1: Simplify the integrand.
We have and .
Also, and .
The integral becomes:
=
=
=
=
=
=

Step 2: Use the substitution t = .
Then dt = .
We need to rewrite the integral in terms of .

=
=
=

Step 3: Use the substitution u = 1 + .
Then du = .
We have .
.

=
=
=
Let u = 1 + . Then du = .
.

=
=
=
=
Therefore, the integral is .

Step 1: Rewrite the integrand.
Let I = .
We can rewrite the integral as: I =

Step 2: Use integration by parts.
Let u = x csc x and dv = .
Then du = csc x - x csc x cot x dx and v = .
Using integration by parts, :
I =
Let's try to find the derivative of .

=
=
=
=

Step 3: Consider the derivative of .
Let f(x) = .
Then f'(x) =
=
=
=
=

Step 4: Consider the derivative of .

Therefore, the integral is .

Step 1: Rewrite the given expression.
Let L = .
L =

Step 2: Take logarithm on both sides.


Step 3: Recognize the Riemann sum.

This is a Riemann sum, so we can write it as an integral:


Step 4: Evaluate the integral using integration by parts.
Let u = and dv = dx.
Then du = and v = x.









Step 5: Compare with ae^b.
L =
So a = 2 and b = .
a + b =
Therefore, a+b = .

Step 1: Use the substitution Let Then, Squaring both sides, Thus, Step 2: Differentiate both sides Using implicit differentiation, Rearranging, Using trigonometric identities, solve the integral, leading to: