Question

Multiple Choice If \(\int_{2}^{5} f(x) d x=18,\) then \(\int_{2}^{5}(f(x)+4) d x=\) \((\mathbf{A}) 20 \quad(\mathbf{B}) 22\) \((\mathrm{C}) 23 \quad(\mathrm{D}) 25 \quad(\mathrm{E}) 30\)

Short answer

The correct answer is (E)30.

Step-by-step solution

Integrate the Given Function

It is given that \(\int_{2}^{5} f(x) d x=18\). This represents the area under the curve of function f(x) from x=2 to x=5. This is the reference area that we have.

Integrate the Incremented Function

The integral \(\int_{2}^{5}(f(x)+4) dx\) represents the area under the curve of function (f(x) + 4) from x=2 to x=5. This is equivalent to \(\int_{2}^{5} f(x) dx + \int_{2}^{5} 4 dx\), which can be calculated as: 18 + 4*(5-2) = 18 + 4*3 = 30. The result is the sum of the area under f(x) (which equals 18) and the area under the constant function 4 (which equals 4 * the length of the interval [2,5] = 4*3 = 12). The total equals 30. Therefore, \(\int_{2}^{5}(f(x)+4) dx = 30\), which corresponds to answer (E).

Key concepts

Integration

Understanding the concept of integration is crucial for mastering calculus. Integration is the mathematical process of finding the integral of a function, which represents the accumulation of values, such as areas under curves, volumes, and other quantities.

Consider it as the opposite operation of differentiation, where we find the slope of a curve; integration, instead, tells us about the space it occupies. In a definite integral, as seen with \( \int_{2}^{5} f(x) dx = 18 \), the boundaries (2 and 5) specify the interval over which we're accumulating the area under the function \( f(x) \). This kind of integral yields a numerical value that represents the total 'accumulated' quantity between the two bounds.

Area Under the Curve

The area under the curve is a fundamental concept tied to definite integration. When we calculate \( \int_{a}^{b} f(x) dx \), we are essentially adding up infinitely many, infinitely thin vertical slices under the curve of \( f(x) \) from \( x=a \) to \( x=b \).

This visualisation is particularly useful for understanding real-world phenomena such as displacement, probability distributions, and work done by a varying force. In the exercise, the original area under \( f(x) \) between 2 and 5 was given as 18, which is an exact figure representing this concept.

Calculus Problems

Calculus problems often require a combination of conceptual understanding and algebraic manipulation. The problem at hand demonstrates a common calculus problem—adding a constant to a function before integrating.

In such problems, recognizing the properties of integrals can lead to a swift solution. Being able to break down a compound function into simpler parts that are easier to integrate separately is a vital skill, as seen in the exercise where \( f(x) \) and 4 were integrated separately before combining the results for the final answer.

Integration Properties

The properties of integration are what make it a powerful tool in mathematics. One such property is linearity, which allows us to add or subtract integrals, as well as to pull constants in and out of the integral sign.

In the provided exercise, this property is exploited to integrate the function \(f(x) + 4\) as a sum of two separate integrals, \( \int_{2}^{5} f(x) dx \) and \( \int_{2}^{5} 4 dx \). This separates the variable part from the constant, making the integration more straightforward and showing that integration is not just a rule-based process but also one that requires strategic thinking.