Find the area of the region bounded by the curve 𝑦 =square root of x^9-9/x^2,

π‘‘β„Žπ‘’ π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠, π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒 π‘₯ = 5

Do you mean

√(x^9-9) / x^2
or
√((x^9-9)/x^2)
or, more probably,
√(x^2-9) / x^2

??
In any case, you have problems at x=0

since y is undefined for x<3 you have a problem. But, using integration by parts,

∫√(x^2-9) / x^2 dx = -arcsin(x/3) - √(x^2-9)/x

now it's even worse. arcsin(x/3) is undefined for x>3 !

Maybe you can figure out what's to be done ...

hello, correction, its √(x^2-9) / x^2

To find the area of the region bounded by the curve 𝑦 = √(π‘₯^9 - 9/π‘₯^2) and the x-axis between π‘₯ = -π‘Ž and π‘₯ = π‘Ž, we can use integration.

Step 1: Find the x-values where the curve intersects the x-axis.
To find these points, we set 𝑦 = 0 and solve for π‘₯:
0 = √(π‘₯^9 - 9/π‘₯^2)
Squaring both sides:
0 = π‘₯^9 - 9/π‘₯^2
Now, we have a polynomial equation that we can solve. However, this equation is quite complex and requires numerical methods to find the solutions. Let's assume one of the solutions is -π‘Ž and the other is π‘Ž.

Step 2: Write the integral for the area.
The area can be calculated by integrating the absolute value of the curve's equation between π‘₯ = -π‘Ž and π‘₯ = π‘Ž. Since the curve 𝑦 = √(π‘₯^9 - 9/π‘₯^2) is above the x-axis, we don't need the absolute value in this case. The integral for the area is:
A = ∫[from -π‘Ž to π‘Ž] 𝑦 dπ‘₯
We need to express 𝑦 in terms of π‘₯ to perform this integral.

Step 3: Express 𝑦 in terms of π‘₯.
𝑦 = √(π‘₯^9 - 9/π‘₯^2)

Step 4: Evaluate the integral.
Now that we have 𝑦 in terms of π‘₯, we can evaluate the integral:
A = ∫[from -π‘Ž to π‘Ž] √(π‘₯^9 - 9/π‘₯^2) dπ‘₯

Unfortunately, step 1 requires numerical methods to find the values of π‘Ž. Once you have those values, you can proceed to step 4 to calculate the area by evaluating the integral.