#### To determine

**To evaluate:**

∫-22(x+3)4-x2dx by writing it as a sum of two integrals and interpret one of those integrals in terms of an area.

#### Answer

6π

#### Explanation

**1) Concept:**

i) The substitution rule:

If u=g(x) is a differentiable function whose range is I and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. Here,g(x) is substituted as u and then g’(x)dx =du.

ii) Indefinite integral

∫xn dx=xn+1n+1+C (n≠-1)

iii) Integrals of the symmetric function

suppose f is continuous on -a, a

If f is odd [f(-x)=-f(x)] function, then ∫-aafxdx=0

**2) Given:**

∫-22(x+3)4-x2dx

3) **Calculation:**

Given that

**∫-22(x+3)4-x2dx**

Multiply and separate the integral term.

**=∫-22x4-x2 +34-x2dx**

=∫-22x4-x2dx+∫-2234-x2dx

Consider the first integral,

∫-22x4-x2dx

Let gx=x 4-x2

Consider,g-x=-x4-x2= -x 4-x2= -g(x)

Since g is an odd function, the first integral is 0.

Now consider second integral,

∫-2234-x2dx

=∫-2234-x2dx

=3∫-224-x2dx

Take,4-x2=y

Squaring both sides

4-x2=y2

Simplify.

x2+y2=4

This is the equation of the circle with a radius 2

The integral represents the area of the top half of a circle multiplied by 3

So, find area of semicircle with radius 2 and multiplied by 3.

Area of semicircle =12πr2

Substitute r=2 in to get 12π(2)2=2π

So required area is 3·2π=6π.

∫-2234-x2dx=3·2π=6π

∫-22x4-x2dx=0

Therefore,

**∫-22(x+3)4-x2dx=6π**

**Conclusion:**

Therefore,

**∫-22(x+3)4-x2dx=6π**