Question

In Problems 1-16, perform the indicated integrations. \(\int_{1}^{2} \frac{d t}{\sqrt{t}+e}\)

Short answer

The integral evaluates to \( 2 \left( \sqrt{2} - e \ln(\sqrt{2} + e) \right) - 2 \left( 1 - e \ln(1 + e) \right).\)

Step-by-step solution

Set up the Integral

The problem requires us to integrate the function \( \frac{1}{\sqrt{t} + e} \) over the interval from 1 to 2 with respect to \( t \). This can be written as \( \int_{1}^{2} \frac{dt}{\sqrt{t} + e} \).

Evaluate the Integral Function

We need to find the antiderivative of \( \frac{1}{\sqrt{t} + e} \). A substitution method can be helpful here. Let \( u = \sqrt{t} \), which implies that \( t = u^2 \) and \( dt = 2u \, du \). So, the integral becomes: \[ \int \frac{2u \, du}{u + e}. \]

Simplify using Substitution

We substitute \( dt = 2u \, du \) into the integral, obtaining: \[ 2 \int \frac{u \, du}{u + e}. \] This expression can be simplified using partial fraction decomposition or direct integration.

Integrate using Direct Integration

The integral \( \int \frac{u}{u+e} \, du \) can be integrated by recognizing it as \( (u + e - e)/(u+e) \), which splits into \( 1 - \frac{e}{u+e} \). Thus, we have: \[ 2 \int (1 - \frac{e}{u+e}) \, du = 2 \left[ u - e \ln|u+e| \right]. \]

Back Substitute and Evaluate Limits

Reverting back to \( t \) by substituting \( u = \sqrt{t} \), we have: \[ 2 \left[ \sqrt{t} - e \ln|\sqrt{t}+e| \right]. \] Evaluate this from \( t = 1 \) to \( t = 2 \).

Compute Definite Integral

Calculate the antiderivative at the boundary points 2 and 1: \[ 2 \times \left( \sqrt{2} - e \ln(\sqrt{2} + e) \right) - 2 \times \left( 1 - e \ln(1 + e) \right). \] Simplify this expression to find the answer.

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, representing the accumulation of quantities over an interval. The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the function being integrated, and \( a \) and \( b \) are the limits of integration. This symbolically expresses the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \). Definite integrals provide a precise way to calculate this area, which is key in many applications, such as determining physical quantities like displacement and work.

In the given exercise, we are evaluating the definite integral \( \int_{1}^{2} \frac{dt}{\sqrt{t} + e} \). Here, the function \( \frac{1}{\sqrt{t} + e} \) is integrated over the interval from 1 to 2. The limits, 1 and 2, define the boundaries within which the area under the curve is calculated.

Calculating a definite integral involves finding the antiderivative of the function and then evaluating it at the upper and lower limits. By substituting these values into the antiderivative, we differentiate between the value at \( b \) and \( a \) to find the exact area.

Substitution Method

The substitution method is a powerful technique for simplifying integrals by changing variables. It is particularly useful when dealing with composite functions where direct integration is complex. The essence of substitution lies in transforming the integration variable to simplify the integrand.

For the exercise, we used the substitution \( u = \sqrt{t} \), implying \( t = u^2 \) and thus \( dt = 2u \, du \). This transforms the integral into \( \int \frac{2u \, du}{u + e} \), which is easier to solve. Substituting can often break down a complicated expression, allowing more straightforward integration techniques.

Choosing an appropriate substitution can significantly reduce the difficulty of calculating an integral. The goal is to convert the integral into a new variable that has a simpler antiderivative, ultimately allowing for easier integration and solution.

Partial Fraction Decomposition

Partial fraction decomposition is used to simplify fractions into a sum of simpler fractions, which can be easier to integrate. However, not all integrals require explicit use of this method. Because the expression in our exercise, \( \frac{u}{u+e} \), can readily be modified into simpler parts for direct integration, partial fraction decomposition was not necessary. But understanding this method is crucial for more complex integrals.

In cases where decomposition is applied:

• The fraction is decomposed into partial fractions.
• Each component is easier to integrate.
• The solutions are then combined to solve the integral completely.

This strategy is particularly helpful when dealing with rational functions, where the denominator can be factored into linear or quadratic terms.

Antiderivatives

An antiderivative is a function whose derivative recovers a given function. Finding an antiderivative is crucial in computing integrals, as the definite integral is essentially the evaluation of an antiderivative at specified limits. The fundamental theorem of calculus links this operation to determining areas under curves.

In this context, we identified the antiderivative of \( \frac{1}{\sqrt{t}+e} \) using substitutions and transformations. The steps led us to the expression \( 2 \left[ \sqrt{t} - e \ln|\sqrt{t}+e| \right] \). This antiderivative was then evaluated at the boundaries of 1 and 2 to find the solved integral value.

Understanding antiderivatives is crucial for solving both definite and indefinite integrals. It allows us to reverse the process of differentiation, thus providing insights into the accumulated quantity over a particular interval, just as in our exercise.