Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int_{0}^{3 / 4} \frac{\sin \sqrt{1-x}}{\sqrt{1-x}} d x\)

Step-by-Step Solution

Answer

The integral evaluates to approximately 0.800.

Step 1: Substitution

To solve the integral \(\int_{0}^{3 / 4} \frac{\sin \sqrt{1-x}}{\sqrt{1-x}} d x\), start by making a substitution. Let \( u = \sqrt{1-x} \), then \( x = 1 - u^2 \). Calculate \( dx \) by differentiating: \( dx = -2u \, du \).

Step 2: Adjust Limits of Integration

When \( x = 0 \), \( u = \sqrt{1-0} = 1 \). When \( x = \frac{3}{4} \), \( u = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \). Therefore, the new limits of integration are from 1 to \( \frac{1}{2} \).

Step 3: Rewrite the Integral

Substitute \( u \) and \( dx \) into the integral:\[\int_{1}^{\frac{1}{2}} \sin(u) \cdot \left(-2u \right) \, du = -2 \int_{1}^{\frac{1}{2}} u \sin(u) \, du\]

Step 4: Change the Limits of Integration

Since we have negative bounds, switch the limits to correspond to the traditional lower to upper bounds and change the sign of the integral:\[2 \int_{\frac{1}{2}}^{1} u \sin(u) \, du\]

Step 5: Integration by Parts

Integrate using integration by parts, where \( dv = \sin(u) \, du \) and \( v = -\cos(u) \). Let \( w = u \) and \( dw = du \). Integration by parts formula is \( \int w \, dv = wv - \int v \, dw \).

Step 6: Apply Integration by Parts

Apply the integration by parts as follows:\[ wv = -u \cos(u) \Bigg|_{\frac{1}{2}}^{1} \]\[- \int v \, dw = \int \cos(u) \, du = \sin(u) \Bigg|_{\frac{1}{2}}^{1} \]Combine these results:\[ 2 \left( -u \cos(u) \bigg|_{\frac{1}{2}}^{1} + \sin(u) \bigg|_{\frac{1}{2}}^{1} \right) \]

Step 7: Evaluate the Boundaries

Evaluate the expression from Step 6 at the boundaries:\[-\left[ 1 \cdot \cos(1) - \frac{1}{2} \cdot \cos\left(\frac{1}{2}\right) \right] + \left[\sin(1) - \sin\left(\frac{1}{2}\right)\right]\]Multiply by 2 and compute the result.

Step 8: Final Calculation

Substitute the values:\[-2(\cos(1) - \frac{1}{2} \cos(\frac{1}{2})) + 2(\sin(1) - \sin(\frac{1}{2}))\]Calculate each term for the final answer.

Key Concepts

Integration by Parts

Integration by parts is a crucial technique in integral calculus for dealing with integrals that are products of functions. The concept is based on the product rule for differentiation and allows us to transform complex integrals into more manageable ones.

The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] where:

• \( u \) is a function that we differentiate.
• \( dv \) is a function that we integrate.
• \( du \) is the derivative of \( u \).
• \( v \) is the integral of \( dv \).

Choosing the right \( u \) and \( dv \) is an art in itself, often requiring practice and insight into function behaviors. Once selected, compute \( du \) and \( v \) and plug them into the formula to simplify the integral.

In our problem, after substitutions, we let \( w = u \) and \( dv = \sin(u) \, du \), resulting in \( v = -\cos(u) \). This choice was strategic to simplify the resulting expression after applying the integration by parts technique.

Substitution Method

The substitution method is a powerful technique used to simplify integrals, particularly when dealing with composite functions. It is often the first step in adjusting the integral to make it more straightforward for other methods like integration by parts or direct integration.

The method involves substituting a part of the integral with a new variable, usually denoted as \( u \).

• The integral limits are also adjusted according to this substitution to reflect the new variable's bounds.
• Once the substitution is made, replace all instances of the original variable with the new one.
• Don't forget to also express \( dx \) in terms of \( du \).

In our problem, we replaced \( x \) by setting \( u = \sqrt{1-x} \), transforming both the bounds and the expression within the integral. This substitution helped in simplifying the integral into a form that was suitable for further techniques like integration by parts.

Definite Integrals

Definite integrals are used to calculate the net area under a curve within specified bounds, from a lower limit to an upper limit. They are a major component of calculus that connect the concept of differentiation with accumulation of quantities.

The definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted as:\[ \int_{a}^{b} f(x) \, dx \] Key characteristics of definite integrals include:

• The result is a concrete numerical value representing the accumulated quantity, not a general function.
• Definite integrals can also be used to evaluate total change over an interval, summing small pieces described by the integral.

In our problem, we transformed limits from \( x \) boundaries to \( u \) to reflect the substitution method's new parameters. After applying integration techniques, we evaluated the integral at these bounds to find the solution. The transition from \( 0 \) to \( \frac{3}{4} \) in \( x \) was converted to \( 1 \) to \( \frac{1}{2} \) in \( u \), emphasizing the need to adjust limits with any substitution.