Definite Integral

To find the value of , we first identify where the function changes form. The expression when . So, we define the absolute value piecewise:

Since the limits of integration are from 0 to 3, we are only concerned with , where .

The integral becomes:

Compute each integral:

Therefore,

Thus, the value of is 9.

1. Integral A: evaluates to by symmetry and properties of definite integrals. Thus, A matches with III.
2. Integral B: is evaluated using symmetry and properties, resulting in . Thus, B matches with IV.
3. Integral C: calculates to 0 as it is an odd function over a symmetric interval. Thus, C matches with II.
4. Integral D: results in by trigonometric identities and integration. Thus, D matches with I.

We start with the integral:

Let , so:

The limits change as follows:

For and for .

Substitute into the integral:

Simplify:

Solve each term: For :

Evaluate:

For :

Combine results:

Final simplification gives:

The given integral is:

Analyze the symmetry of the integral. The limits of the integral are symmetric about , and the integrand contains terms that involve trigonometric functions , , , and . Specifically, consider the property:

For the integrand:

Using the trigonometric substitutions:

we find that the integrand satisfies the property:

Since is odd with respect to , the integral over the symmetric interval evaluates to 0.

Thus:

Let:

I = .

Substitute , so or . The limits become:

,

.

The integral becomes:

I = .

Split the fraction:

.

Thus:

I = .

Integrate:

I = .

Simplify:

I = .

Combine terms:

I = .

Simplify further:

I = .

Simplify:

I = .

This simplifies to:

I = .

Thus:

.