Problem 23 - Calculus

Question

If \(\int_{0}^{1} f(x) d x=4, \int_{0}^{2} f(x) d x=2,\) and \(\int_{0}^{2} g(x) d x=\) -3 , evaluate each integral. (a) \(\int_{1}^{2} f(x) d x\) (b) \(\int_{1}^{0} f(x) d x\) (c) \(\int_{0}^{2} 3 f(u) d u\) (d) \(\int_{0}^{2}[2 g(x)-3 f(x)] d x\) (e) \(\int_{0}^{-2} f(-x) d x\)

Step-by-Step Solution

Answer

(a) -2, (b) -4, (c) 6, (d) -12, (e) 2.

Step 1: Use the Property of Definite Integrals for (a)

To find \( \int_{1}^{2} f(x) \, dx \), we use the property of definite integrals. Given \( \int_{0}^{2} f(x) \, dx = 2 \) and \( \int_{0}^{1} f(x) \, dx = 4 \), we know that \( \int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx \). So, \( \int_{1}^{2} f(x) \, dx = 2 - 4 = -2 \).

Step 2: Reverse Limits for (b)

To evaluate \( \int_{1}^{0} f(x) \, dx \), we use the property that reversing the limits of integration changes the sign of the integral. Therefore, \( \int_{1}^{0} f(x) \, dx = -\int_{0}^{1} f(x) \, dx = -4 \).

Step 3: Factor the Constant for (c)

For \( \int_{0}^{2} 3 f(u) \, du \), you can factor out the constant: \( 3 \int_{0}^{2} f(u) \, du \). We know \( \int_{0}^{2} f(u) \, du = 2 \), hence \( 3 \times 2 = 6 \).

Step 4: Apply Linear Combination for (d)

To solve \( \int_{0}^{2}[2 g(x) - 3 f(x)] \, dx \), apply the linear combination rule: \( 2 \int_{0}^{2} g(x) \, dx - 3 \int_{0}^{2} f(x) \, dx \). From the given, \( \int_{0}^{2} g(x) \, dx = -3 \) and \( \int_{0}^{2} f(x) \, dx = 2 \). Compute: \( 2(-3) - 3(2) = -6 - 6 = -12 \).

Step 5: Shift and Reverse Limits for (e)

For \( \int_{0}^{-2} f(-x) \, dx \), perform a substitution: let \( u = -x \), giving \( du = -dx \). Change limits: when \( x = 0, u = 0 \) and when \( x = -2, u = 2 \). Therefore, \( \int_{0}^{-2} f(-x) \, dx = \int_{0}^{2} f(u) (-du) = -\int_{2}^{0} f(u) \, du = \int_{0}^{2} f(u) \, du \) due to reversing limits. Therefore, the value is \( 2 \).

Key Concepts

Definite Integrals

In calculus, the definite integral provides a way to calculate the accumulated quantity of a function over a specific interval. It is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper bounds, respectively. The definite integral measures the net area between the function \( f(x) \) and the \( x \)-axis from \( x = a \) to \( x = b \).
This area can be either positive or negative depending on whether the function is above or below the \( x \)-axis within the interval. To compute the definite integral, the signed areas under the curve from each sub-interval are summed. It results in a scalar value reflecting the total accumulation of \( f(x) \) across the domain of integration, giving insight into the function's behavior over that range.

Integral Properties

Several properties of definite integrals simplify calculations and broaden the range of applications. The {} - Additivity Property: If you have a total interval from \( a \) to \( c \), you can split it at some intermediate point \( b \) as: \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \).
- Reversal Property: Flipping the limits of the integral changes the sign: \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).
- Zero Width Interval: If the bounds are the same (\( a = b \)), the integral is zero: \( \int_{a}^{a} f(x) \, dx = 0 \).
Understanding these properties allows us to manipulate and solve various integral problems with more flexibility, improving problem-solving efficiency.

Linear Combination of Integrals

A fascinating property in calculus is the linear combination of integrals. This allows us to manage multiple integrals when functions are combined linearly. For example, if you have coefficients \( c \) and \( d \), then
\[ \int_{a}^{b} [c \cdot f(x) + d \cdot g(x)] \, dx = c \cdot \int_{a}^{b} f(x) \, dx + d \cdot \int_{a}^{b} g(x) \, dx \]
This characteristic simplifies the integration of expressions where multiple functions are involved. It permits integration to be done separately for each function and their individual results to be adjusted by their respective coefficients and summed. This immensely helps in breaking down more complex integrals into manageable parts. For instance, in the original problem, we used this approach by computing each component integral separately from the composite expression for ease and accuracy.

U-Substitution in Integrals

U-substitution is a technique used to simplify integrals by performing a variable change. It alters the variables and limits, allowing for easier integration. In core essence, it is akin to reverse application of the chain rule for derivatives. To use u-substitution,
- Choose \( u \) as a function of \( x \) such that the integral becomes simpler.
- Compute \( du \), which translates \( dx \) in the integral.
- Change the limits of integration according to the new variable \( u \).
- Substitute \( u \) and \( du \) back into the integral to complete the process.
In our exercise, we applied this with \( u = -x \), translating the integral's limits and reversing the ordered bounds due to negative differential. Mastering u-substitution enriches the toolbox for handling integrals involving variable-dependent transformations.