Questions: ∫ from -1 to 1 (x^2 + 3x - 1/x^2) dx

Transcript text: $\int_{-1}^{1}\left(x^{2}+3 x-\frac{1}{x^{2}}\right) d x$

Solution

Solution Steps

To solve the given integral, we will use Python's scipy.integrate library which provides a function quad to compute definite integrals. We will define the integrand function and then use quad to evaluate the integral from -1 to 1.

Step 1: Break Down the Integral

We are given the integral: \[ \int_{-1}^{1}\left(x^{2}+3x-\frac{1}{x^{2}}\right) dx \]

Step 2: Analyze the Integrand

Notice that the integrand $ x^2 + 3x - \frac{1}{x^2} $ is composed of three separate terms. We can split the integral into three separate integrals: \[ \int_{-1}^{1} x^2 \, dx + \int_{-1}^{1} 3x \, dx - \int_{-1}^{1} \frac{1}{x^2} \, dx \]

Step 3: Evaluate Each Integral Separately

Integral 1: $\int_{-1}^{1} x^2 \, dx$

Since $ x^2 $ is an even function, we can use the property of definite integrals for even functions over symmetric limits: \[ \int_{-1}^{1} x^2 \, dx = 2 \int_{0}^{1} x^2 \, dx \] \[ = 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \] \[ = 2 \cdot \frac{1}{3} = \frac{2}{3} \]

Integral 2: $\int_{-1}^{1} 3x \, dx$

Since $ 3x $ is an odd function, the integral of an odd function over symmetric limits is zero: \[ \int_{-1}^{1} 3x \, dx = 0 \]

Integral 3: $\int_{-1}^{1} \frac{1}{x^2} \, dx$

The function $ \frac{1}{x^2} $ is undefined at $ x = 0 $, so the integral does not converge. Therefore, this integral is improper and cannot be evaluated in the standard sense.

Final Answer

Since the third integral does not converge, the entire integral does not converge. Therefore, the integral is undefined.

\[ \boxed{\text{undefined}} \]

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