Question

Evaluate the integrals. If the integral diverges, answer "diverges."\(\int_{0}^{1} \frac{d x}{\sqrt{1-x}}\)

Short answer

The integral evaluates to 2.

Step-by-step solution

Identify the Type of Integral

The integral given is \( \int_{0}^{1} \frac{dx}{\sqrt{1-x}} \). This is a definite integral over the interval [0,1]. Notice that the integrand has a singularity at \(x = 1\) since \(\sqrt{1-x} = 0\) when \(x = 1\). This means the integral could potentially diverge at this endpoint.

Evaluate the Antiderivative

The antiderivative of \(\frac{1}{\sqrt{1-x}}\) can be found as follows. Use the substitution \( u = 1-x \), then \( du = -dx \) or \( dx = -du \). This converts the integral to \(-\int \frac{du}{\sqrt{u}}\), whose antiderivative is \(-2\sqrt{u} + C\). Substituting back \(u = 1-x\), the antiderivative is \(-2\sqrt{1-x} + C\).

Apply the Fundamental Theorem of Calculus

Using the antiderivative \(-2\sqrt{1-x}\), apply the Fundamental Theorem of Calculus:\[\lim_{b \to 1^-} \left[-2\sqrt{1-x}\right]_{0}^{b} = \lim_{b \to 1^-} \left[-2\sqrt{1-b} + 2\sqrt{1-0}\right] = \lim_{b \to 1^-} [-2\sqrt{1-b} + 2]\]

Evaluate the Limit

Evaluate the limit: \[-e^{u} = \lim_{b \to 1^-} [2 - 2\sqrt{1-b}] = 2 - \lim_{b \to 1^-} 2\sqrt{1-b.\]As \( b \to 1^- \), \( \sqrt{1-b} \to 0 \). Therefore, the limit becomes \(2 - 0 = 2\).

Conclusion

The integral evaluates to a finite value of 2, meaning it converges.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a hugely important concept that connects differentiation and integration. It comes in two parts and provides a way to evaluate definite integrals with ease.
The theorem can be summarized as follows:

• The first part states that if a function is continuous over an interval, then it has an antiderivative over that interval, which means there is a function whose derivative gives you the original function.
• The second part gives the practical application, allowing us to compute a definite integral by subtracting the values of an antiderivative at the boundary points of the interval.

In our original exercise, we used the Fundamental Theorem of Calculus to evaluate the integral \( \int_{0}^{1} \frac{dx}{\sqrt{1-x}} \) by first finding its antiderivative and then calculating the difference between the upper and lower limits of integration. This is what allowed us to determine whether the integral converged to a finite value or not.

Antiderivatives

An antiderivative of a function is another function whose derivative is the original function. Finding antiderivatives is often part of solving definite integrals.
They are also known as "indefinite integrals."
This process often involves techniques such as substitution, which is what we used in the original exercise.

Substitution Method

Substitution is used to simplify a given integral into a form that is easier to evaluate. In the exercise, we set \( u = 1-x \) which simplifies the integrand \( \frac{1}{\sqrt{1-x}} \) to \( \frac{1}{\sqrt{u}} \). This allowed us to find the antiderivative more easily using known integral formulas.

Importance of Constants

When finding antiderivatives, remember the constant \(+ C\). However, in definite integrals, these constants cancel out, simplifying the overall computation. The antiderivative is crucial for applying the Fundamental Theorem of Calculus to find the exact value of definite integrals.

Convergence and Divergence of Integrals

When evaluating integrals, especially improper ones, determining convergence or divergence is essential.Convergence means that the integral equates to a finite number, while divergence means it does not settle on any particular value.

Improper Integrals

Improper integrals can have infinite limits or points of discontinuity in the interval.
In the problem \( \int_{0}^{1} \frac{dx}{\sqrt{1-x}} \), the function is undefined, or singular, at \( x = 1 \). However, in this case, we found the limit while approaching 1 from the left side, which determined that it converged.

The Role of Limits

The calculations involving limits, where the function might be undefined, are crucial to distinction between convergence and divergence.These limits help reveal whether the areas under the function approach a finite value or grow indefinitely."