Question

In Problems 1-16, perform the indicated integrations. \(\int_{0}^{1} \frac{\sqrt{t}}{t+1} d t\)

Short answer

The integral evaluates to \( 2 - \frac{\pi}{2} \).

Step-by-step solution

Understand the Integral Problem

We are given the definite integral \( \int_{0}^{1} \frac{\sqrt{t}}{t+1} \, dt \). This means we need to find the area under the curve of the function \( f(t) = \frac{\sqrt{t}}{t+1} \) from \( t=0 \) to \( t=1 \). Since the integrand is a combination of a square root and a rational function, substitution might make the integration easier.

Choose a Suitable Substitution

Substitute \( t = u^2 \). Then, \( dt = 2u \, du \). The limits of integration change as follows: when \( t = 0 \), \( u = 0 \), and when \( t = 1 \), \( u = 1 \). Thus, the integral becomes \( \int_{0}^{1} \frac{u}{u^2 + 1} \cdot 2u \, du \), which simplifies to \( 2 \int_{0}^{1} \frac{u^2}{u^2 + 1} \, du \).

Simplify the Integrand

The integrand \( \frac{u^2}{u^2 + 1} \) can be rewritten using algebra: \( \frac{u^2}{u^2 + 1} = 1 - \frac{1}{u^2 + 1} \). Thus, the integral becomes \( 2 \int_{0}^{1} \left( 1 - \frac{1}{u^2 + 1} \right) \, du \).

Integrate the Simplified Expressions

Now integrate the expression: \( 2 \int_{0}^{1} \left( 1 - \frac{1}{u^2 + 1} \right) \, du = 2 \left[ \int_{0}^{1} 1 \, du - \int_{0}^{1} \frac{1}{u^2 + 1} \, du \right] \). Calculate the integrals: \( \int_{0}^{1} 1 \, du = \left[ u \right]_{0}^{1} = 1 \) and \( \int_{0}^{1} \frac{1}{u^2 + 1} \, du = \left[ \tan^{-1}(u) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \).

Evaluate and Finalize the Integral

Combine the results of the integrals: \( 2 \times \left[ 1 - \frac{\pi}{4} \right] = 2 - \frac{\pi}{2} \). This gives the final evaluation of the original integral.

Key concepts

Integration by Substitution

When faced with complex integrals combining functions like square roots and rational expressions, integration by substitution can be a powerful tool. This technique simplifies integration by converting the variable of integration into a new variable, thus transforming the integral into a simpler form.

In our problem, we start with the integral \( \int_{0}^{1} \frac{\sqrt{t}}{t+1} \, dt \). Recognizing that the presence of \( \sqrt{t} \) can complicate the process, we choose a substitution that will eliminate this complexity. This is achieved by using \( t = u^2 \) which transforms \( dt \) into \( 2u \, du \).

This strategic choice converts the integral into a form involving only rational expressions, \( \int_{0}^{1} \frac{u \cdot 2u}{u^2 + 1} \, du \), further simplifying to \( 2 \int_{0}^{1} \frac{u^2}{u^2 + 1} \, du \). Thus, substitution makes solving this integral much more manageable.

Limits of Integration

The process of integration with substitution not only involves changing the variable of integration but also requires adjusting the limits of integration accordingly.

In this exercise, when we substitute \( t = u^2 \), we must also update the limits of integration from the original bounds on \( t \). Initially, our limits were from \( t = 0 \) to \( t = 1 \).

By substituting \( t = u^2 \), the new limits transform as follows:

• When \( t = 0 \), we solve for \( u \) giving \( u = \sqrt{0} = 0 \).
• When \( t = 1 \), we solve for \( u \) giving \( u = \sqrt{1} = 1 \).

Without adapting these limits correctly, the integration becomes incorrect and the results unusable.

Algebraic Manipulation

Algebraic manipulation is a crucial step in dealing with integrals, especially when complex expressions are involved.

In our revised integral \( 2 \int_{0}^{1} \frac{u^2}{u^2 + 1} \, du \), simplifying the integrand allows us to split it into simpler parts. By expressing it as \( 1 - \frac{1}{u^2 + 1} \), we effectively break the integral into two parts:

• \( 2 \int_{0}^{1} 1 \, du \), which simplifies directly as the integral of a constant.
• \( 2 \int_{0}^{1} \frac{1}{u^2 + 1} \, du \), which points toward logarithmic or inverse trigonometric identities.

This clever simplification is pivotal for performing the integration separately and easily.

Inverse Trigonometric Functions

Inverse trigonometric functions often appear in integrals involving quadratic expressions. Understanding these functions can provide elegant solutions.

In this problem, the expression \( \frac{1}{u^2 + 1} \) is recognized as the derivative of \( \tan^{-1}(u) \).

This places the problem into the context of inverse trigonometric functions, allowing for a straightforward antiderivative:

• \( \int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) + C \)

Calculating from the updated limits, we find:

• \( \left[ \tan^{-1}(u) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 \)

Using inverse trig functions can transform a complicated integral into a straightforward task.