Problem 42 - Calculus Early Transcendentals

Question

Suppose \(\int_{1}^{4} f(x) d x=8\) and \(\int_{1}^{6} f(x) d x=5 .\) Evaluate the following integrals. a. \(\int_{1}^{4}(-3 f(x)) d x \) b. \( \int_{1}^{4} 3 f(x) d x\) c. \(\int_{6}^{4} 12 f(x) d x\) d. \(\int_{4}^{6} 3 f(x) d x\)

Step-by-Step Solution

Answer

#Answer: a) $\int_{1}^{4}(-3 f(x)) d x = -24$ b) $\int_{1}^{4}(3 f(x)) d x = 24$ c) $\int_{6}^{4} 12 f(x) d x = 36$ d) $\int_{4}^{6}(3 f(x)) d x = -9$

Step 1: Identify given integrals

We have two given integrals: 1. \(\int_{1}^{4} f(x) d x = 8\) 2. \(\int_{1}^{6} f(x) d x = 5\) Now let's solve each part of the exercise.

Step 2: Evaluating the integrals in part a

We want to evaluate \(\int_{1}^{4}(-3 f(x)) d x\). Use the linearity of the integral: $$\int_{1}^{4}(-3 f(x)) d x = -3 \int_{1}^{4} f(x) d x$$ Now substitute the given value of the integral of \(f(x)\) from \(1\) to \(4\): $$-3 \int_{1}^{4} f(x) d x = -3 (8) = -24$$ So, \(\int_{1}^{4}(-3 f(x)) d x = -24\).

Step 3: Evaluating the integrals in part b

We want to evaluate \(\int_{1}^{4} 3 f(x) d x\). Use the linearity of the integral: $$\int_{1}^{4}(3 f(x)) d x = 3 \int_{1}^{4} f(x) d x$$ Now substitute the given value of the integral of \(f(x)\) from \(1\) to \(4\): $$3 \int_{1}^{4} f(x) d x = 3 (8) = 24$$ So, \(\int_{1}^{4}(3 f(x)) d x = 24\).

Step 4: Evaluating the integrals in part c

We want to evaluate \(\int_{6}^{4} 12 f(x) d x\). First reverse the bounds of integration: $$\int_{6}^{4} 12 f(x) d x = - \int_{4}^{6} 12 f(x) d x$$ Next, use the linearity of the integral: $$- \int_{4}^{6} 12 f(x) d x = -12 \int_{4}^{6} f(x) d x$$ Now use the property that we can split the integral into smaller intervals: $$-12 \int_{4}^{6} f(x) d x = -12 \left( \int_{1}^{6} f(x) d x - \int_{1}^{4} f(x) d x \right)$$ Now substitute the given values of the integrals of \(f(x)\) and simplify: $$-12 \left( 5 - 8 \right) = -12(-3) = 36$$ So, \(\int_{6}^{4} 12 f(x) d x = 36\).

Step 5: Evaluating the integrals in part d

We want to evaluate \(\int_{4}^{6} 3 f(x) d x\). Use the linearity of the integral: $$\int_{4}^{6}(3 f(x)) d x = 3 \int_{4}^{6} f(x) d x$$ Now use the property that we can split the integral into smaller intervals: $$3 \int_{4}^{6} f(x) d x = 3 \left( \int_{1}^{6} f(x) d x - \int_{1}^{4} f(x) d x \right)$$ Now substitute the given values of the integrals of \(f(x)\) and simplify: $$3 \left( 5 - 8 \right) = 3(-3) = -9$$ So, \(\int_{4}^{6}(3 f(x)) d x = -9\).

Key Concepts

Linearity of Integrals

The linearity of integrals is a powerful property that helps in simplifying complex integrals. This property states that integrals can be distributed across addition and can also be factored out of a constant multiplier. In simple terms:

• If you have an integral of a function multiplied by a constant, you can move the constant outside the integral. For example, \( \int a \times f(x) \, dx = a \int f(x) \, dx \).
• Similarly, the integral of a sum of functions is the sum of their integrals: \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \).

This property simplifies calculations significantly. In the current problem, we applied linearity to factor constants from the integrals, allowing us to easily calculate results such as \( -3 \int_{1}^{4} f(x) \, dx \) and \( 3 \int_{1}^{4} f(x) \, dx \). By moving the constant out, calculations became a simple multiplication rather than complex integration calculations.

Reversing Bounds

In integration, the limits or bounds of the integral indicate the start and end points of integration. The order of these bounds matters and reversing them changes the sign of the integral. If you have an integral from a higher bound to a lower bound, you can rewrite it with reversed bounds by introducing a negative sign. Simply, \( \int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx \). This concept helps manage integrations in different scenarios.
For part c in the exercise, we encountered \( \int_{6}^{4} 12 f(x) \, dx \). By reversing the bounds to become \( \int_{4}^{6} 12 f(x) \, dx \) with a minus sign, we streamline the integration process, aligning it with the known integrals from lower to higher limits. This adjustment enabled smooth progression towards finding the desired value.

Interval Splitting

Interval splitting, or breaking down an integral over multiple intervals, is used when you have information about integrals over smaller intervals but need to evaluate larger sections. It involves expressing one integral as a sum or difference of other integrals across subdivisions of the domain. Mathematically, \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \).
In the provided exercise, we used interval splitting in parts c and d. We didn't directly have the value of \( \int_{4}^{6} f(x) \, dx \), but by using the given values \( \int_{1}^{6} f(x) \, dx \) and \( \int_{1}^{4} f(x) \, dx \), we calculated it as the difference between the two known integrals. This approach allowed us to derive a solution that would otherwise be difficult to compute without the splitting method.

Integral Evaluation Steps

Evaluating integrals requires a sequence of systematic steps, particularly when dealing with known integral values. Here's a basic overview of this method:

• Identify the given integral values and labels to know what you can work with.
• Use properties such as linearity and reversing bounds to simplify integration tasks.
• Apply interval splitting, if necessary, to evaluate parts of the integral not directly covered by the given information.
• Finally, calculate the desired integral using simplified expressions and substitute known values to get the result.

For instance, when solving part a, we utilized linearity to move the scalar -3 outside of the integral \( \int_{1}^{4}(-3 f(x)) \, dx \). With the simplified form, it was straightforward to apply the given information to arrive at the value. Following these logical steps helps in solving integral problems efficiently and accurately.