Question

Properties of integrals Suppose \(f(x) \geq 0\) on [0,2] \(f(x) \leq 0\) on \([2,5], \int_{0}^{2} f(x) d x=6,\) and \(\int_{2}^{5} f(x) d x=-8\) Evaluate the following integrals. a. \(\int_{0}^{5} f(x) d x\) b. \(\int_{0}^{5}|f(x)| d x\) c. \(\int_{2}^{5} 4|f(x)| d x\) d. \(\int_{0}^{5}(f(x)+|f(x)|) d x\)

Short answer

a. Evaluate the integral \(\int_{0}^{5} f(x) d x\). The integral of f(x) from 0 to 5 can be found by summing the integrals from 0 to 2 and from 2 to 5: \(\int_{0}^{5} f(x) dx = \int_{0}^{2} f(x) dx + \int_{2}^{5} f(x) dx = 6 + (-8) = -2\) So, \(\int_{0}^{5} f(x) dx = -2\). b. Evaluate the integral \(\int_{0}^{5} |f(x)| d x\). To find the integral of the absolute value of f(x) from 0 to 5, we divide the integral into two intervals and replace |f(x)| with f(x) and -f(x) in the respective intervals: \(\int_{0}^{5} |f(x)| dx = \int_{0}^{2} f(x) dx - \int_{2}^{5} f(x) dx = 6 - (-8) = 14\) So, \(\int_{0}^{5} |f(x)| dx = 14\). c. Evaluate the integral \(\int_{2}^{5} 4|f(x)| d x\). To find the integral of 4 times the absolute value of f(x) from 2 to 5, we replace 4|f(x)| with -4f(x) in the given interval: \(\int_{2}^{5} 4|f(x)| dx = -4 \int_{2}^{5} f(x) dx = -4 (-8) = 32\) So, \(\int_{2}^{5} 4|f(x)| dx = 32\). d. Evaluate the integral \(\int_{0}^{5}(f(x)+|f(x)|) d x\). To find the integral of (f(x)+|f(x)|) from 0 to 5, we divide the integral into two intervals and replace (f(x)+|f(x)|) with 2f(x) and 0 in the respective intervals: \(\int_{0}^{5}(f(x)+|f(x)|) dx = \int_{0}^{2}(2 f(x)) dx + \int_{2}^{5}(0) dx = 2 \int_{0}^{2} f(x) dx + 0 = 2(6) = 12\) So, \(\int_{0}^{5}(f(x)+|f(x)|) dx = 12\).

Step-by-step solution

Divide the integral of f(x) over the given range into two intervals:

$$\int_{0}^{5} f(x) dx = \int_{0}^{2} f(x) dx + \int_{2}^{5} f(x) dx$$

Use the given values to solve the integral

$$\int_{0}^{5} f(x) dx = 6 + (-8)$$

Calculate the result

$$\boxed{\int_{0}^{5} f(x) dx = -2}$$ b. Evaluate the integral \(\int_{0}^{5} |f(x)| d x\)

Divide the integral of |f(x)| over the given range into two intervals:

$$\int_{0}^{5} |f(x)| dx = \int_{0}^{2} |f(x)| dx + \int_{2}^{5} |f(x)| dx$$

Replace |f(x)| with f(x) and -f(x) in the respective intervals

$$\int_{0}^{5} |f(x)| dx = \int_{0}^{2} f(x) dx - \int_{2}^{5} f(x) dx$$

Use the given values to solve the integral

$$\int_{0}^{5} |f(x)| dx = 6 - (-8)$$

Calculate the result

$$\boxed{\int_{0}^{5} |f(x)| dx = 14}$$ c. Evaluate the integral \(\int_{2}^{5} 4|f(x)| d x\)

Replace 4|f(x)| with -4f(x) in the given interval

$$\int_{2}^{5} 4|f(x)| dx = -4 \int_{2}^{5} f(x) dx$$

Use the given values to solve the integral

$$\int_{2}^{5} 4|f(x)| dx = -4 (-8)$$

Calculate the result

$$\boxed{\int_{2}^{5} 4|f(x)| dx = 32}$$ d. Evaluate the integral \(\int_{0}^{5}(f(x)+|f(x)|) d x\)

Divide the integral of (f(x)+|f(x)|) over the given range into two intervals:

$$\int_{0}^{5}(f(x)+|f(x)|) dx = \int_{0}^{2}(f(x)+|f(x)|) dx + \int_{2}^{5}(f(x)+|f(x)|) dx$$

Replace f(x)+|f(x)| with the appropriate expressions in the respective intervals

$$\int_{0}^{5}(f(x)+|f(x)|) dx = \int_{0}^{2}(2 f(x)) dx + \int_{2}^{5}(0) dx$$

Solve the integral in each interval

$$\int_{0}^{5}(f(x)+|f(x)|) dx = 2 \int_{0}^{2} f(x) dx + 0$$

Use the given values to solve the integral

$$\int_{0}^{5}(f(x)+|f(x)|) dx = 2(6)$$

Calculate the result

$$\boxed{\int_{0}^{5}(f(x)+|f(x)|) dx = 12}$$

Key concepts

Definite Integral

Understanding the concept of a definite integral is crucial for solving many problems in calculus. A definite integral represents the accumulation of quantities, such as the area under a curve, between two specified values on the x-axis, known as the limits of integration.

Consider our exercise, which requires evaluating the integral of a function, f(x), over a certain interval. In mathematical terms, the definite integral of f(x) from a to b is represented as \[\int_{a}^{b} f(x) dx\]. This means we are summing up all the infinitesimally small areas under the curve f(x) from point a to point b.

The properties of integrals allow us to split and combine integrals over adjacent intervals, which comes in handy, especially when we know values for subintervals. For example, given \[\int_{0}^{2} f(x) dx = 6\] and \[\int_{2}^{5} f(x) dx = -8\], we can find \[\int_{0}^{5} f(x) dx\] by simply adding these two values together: \[\int_{0}^{5} f(x) dx = \int_{0}^{2} f(x) dx + \int_{2}^{5} f(x) dx = 6 + (-8) = -2\].

This result gives us not just the area, but the net area, considering both the positive and negative regions under the curve.

Integral Evaluation

Integral evaluation involves finding the value of an integral, which is typically done by applying anti-differentiation techniques or using known integral values. To solve an integral, we often reconstruct its original function from its derivative - a process known as finding the antiderivative.

In cases like our exercise, when given definite integrals and specific function properties, we can directly insert the values without finding the antiderivative. For instance, the integral \[\int_{0}^{5} f(x) dx\] is provided as a sum of two other known integrals. Knowing that integral evaluation is linear, these types of problems become much simpler to handle.

For the absolute value integrals, evaluation can become a bit trickier. The absolute value function, which is always non-negative, changes our approach. In the case of \[\int_{0}^{5}|f(x)| dx\], the integral on the interval where f(x) is negative is flipped in sign, converting it into a positive area. Evaluating integral parts with absolute value requires careful attention to the function's behavior on the given interval.

Absolute Value Integrals

Working with absolute value integrals may present additional challenges, but understanding the core idea simplifies the process. The absolute value guarantees that the integrand, the function being integrated, is always nonnegative, which reflects the total area between the curve and the x-axis, regardless of whether the function dips below the axis.

In our exercise, we are asked to evaluate \[\int_{0}^{5}|f(x)| dx\] and \[\int_{2}^{5} 4|f(x)| dx\]. For each interval where f(x) has a consistent sign, we can simplify the absolute value. When f(x) is positive on [0,2], |f(x)| becomes simply f(x), and where f(x) is negative on [2,5], |f(x)| becomes -f(x).

Moreover, when evaluating the integral of a constant multiplied by the absolute value of a function, we can factor out the constant, as seen with \[\int_{2}^{5} 4|f(x)| dx\]. The constant 4 can be positioned outside the integral, facilitating our calculation. Ignoring these properties would lead to errors, so it's essential to consider the absolute values on the specific segments of the function's domain when evaluating these types of integrals.