Integral Calculus

We are asked to find the value of the integral .

We will solve this using substitution. Let:

,

so that , or equivalently, .

Now, the limits of integration change accordingly. When , , and when , .

The integral becomes:

.

Now, we can integrate , which gives . So the integral becomes:

.

This simplifies to:

.

The correct answer is .

We are asked to find the area of the region bounded by the curves and .

To find the area, we first need to determine the points of intersection of the two curves. Set :

implies , which simplifies to . This gives the solutions and .

Thus, the curves intersect at and . We will compute the area between the curves from to .

The area between the curves is given by the integral:

.

Now, compute the integral:

,

and

.

Thus, the area is:

.

The correct answer is .

Given:

take,

⇢⇢[by dividing on both numerator and denominator]

take , then derivate both sides w.r.t x we get

Bye applying this we can write ,

Given:

So to solve this question, we can again use partial fraction decomposition.

Step 1:

Factorize the denominator. can be factored as

Step 2:

Partial fraction decomposition. The expression can be rewritten as the sum of two fractions with unknown constants and :

Step 3:

Now to find the values of A and B, we need to find a common denominator, which is , and then equate the numerators:

Now, solve for A and B by comparing coefficients:

(by comparing the coefficients of ) ⇢(by comparing the constant terms)

Adding the two equations:

⇒

Substituting the value of one of the equations to find we get

Step 4:

Now we can re-write the parent expression as,

(Ans.)

Step 1: Use the power-reduction identity for :

Step 2: Rewrite the integral:

Step 3: Separate the integral:

Step 4: Integrate each term:

Step 5: Simplify the expression:

Conclusion: The correct answer is .

The integral we need to evaluate is:

Let . Then , and .

When , .

When , .

The integral becomes:

We are given the integral: To simplify this integral, let us perform the substitution .
Therefore: Now, the limits of integration change with the substitution. When , we get . When , we get . Thus, the integral becomes: This is a standard form of a trigonometric integral, and after evaluating the integral (using known integrals or a suitable technique), we get: Thus, the value of the integral is .
Thus, the correct answer is option (D), .

We are given the integral: Notice that the integrand is the product of two polynomials.
We can simplify the multiplication first. Expand the terms: First, distribute with each term of : Now, collect like terms: Now, observe that this expression can be simplified further, but we notice the form of the answer choices. Since the integral involves a perfect square and matches the pattern of the answer choices, we recognize that: Thus, the integral simplifies to the form given in option (C).
Thus, the correct answer is option (C), .