Question

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. \(I=\int_{0}^{1}\left(x^{3}-2 x\right) d x=-\frac{3}{4}\) a. \(\int_{0}^{1}\left(4 x-2 x^{3}\right) d x\) b. \(\int_{1}^{0}\left(2 x-x^{3}\right) d x\)

Short answer

Question: What are the values of the integrals: a. \(\int_{0}^{1}\left(4 x-2 x^{3}\right) d x\) b. \(\int_{1}^{0}\left(2 x-x^{3}\right) d x\) Solution: a. The value of the integral \(\int_{0}^{1}\left(4 x-2 x^{3}\right) d x\) is \(\frac{3}{2}\). b. The value of the integral \(\int_{1}^{0}\left(2 x-x^{3}\right) d x\) is \(\frac{3}{4}\).

Step-by-step solution

Identifying Properties of Integrals to Apply

We can notice that the integrand in part a \(\left(4 x - 2x^{3}\right)\) is simply the negative of the given integrand \(\left(x^{3}-2 x\right)\) multiplied by a constant (2). We can apply the property of integrals: \(\int af(x)dx = a\int f(x) dx\) and manipulate the given integral.

Apply the constant factor property

We factor out the constant (2) and rewrite the integrand of part a as follows: \(\int_{0}^{1}\left(4 x-2 x^{3}\right) d x = 2\int_{0}^{1}\left(- x^{3} + 2x\right) d x\) Now, we can observe that the integrand \((- x^{3} + 2x)\) is the negative version of the integrand \(\left(x^3 - 2x\right)\) in the given integral \(I\). Therefore, we can rewrite the integral as: \(\int_{0}^{1}\left(4 x-2 x^{3}\right) d x = 2\left( -\int_{0}^{1}\left(x^{3}-2 x\right) d x\right)\) Since we know the value of the given integral \(I\), we can now substitute it into the equation: \(\int_{0}^{1}\left(4 x-2 x^{3}\right) d x = 2\left( -(-\frac{3}{4})\right)\)

Solve to obtain the value of the integral

Evaluate the expression: \(\int_{0}^{1}\left(4 x-2 x^{3}\right) d x = 2\left(\frac{3}{4}\right) = \frac{3}{2}\) Now let's find the value of the second integral in part b: b. \(\int_{1}^{0}\left(2 x-x^{3}\right) d x\)

Identifying Properties of Integrals to Apply

We can notice that the limits of integration are reversed compared to the given integral \(I\). We can apply the property of integrals that if \(a>b\), then \(\int_a^b f(x) dx = - \int_b^a f(x)dx\). Also, the integrand in part b \(\left(2x - x^{3}\right)\) is a linear combination of the terms in the given integrand \(\left(x^{3}-2 x\right)\). We will apply these properties to manipulate the integral.

Apply the Reversed Limits Property

Using the property of reversing limits, we can rewrite the integral: \(\int_{1}^{0}\left(2 x-x^{3}\right) d x = -\int_{0}^{1}\left(2 x- x^{3}\right) d x\)

Rewrite the integrand as a linear combination of the given integral

We can rewrite the integrand of part b as a linear combination of the terms in the given integral: \(-\int_{0}^{1}\left(2 x- x^{3}\right) d x= -\left(\int_{0}^{1} 2xdx - \int_{0}^{1} x^{3} dx\right)\) Notice that the two integrals inside the parentheses corresponds to portions of the given integral \(I\). In fact, the expression inside the parentheses can be rewritten as: \(2\int_{0}^{1} (- x)dx + \int_{0}^{1} x^{3} dx = -2I + I = -I\) so, substituting the value of \(I\) obtained from the given integral: \(\int_{1}^{0}\left(2 x-x^{3}\right) d x= -I = -\left(-\frac{3}{4}\right)\)

Solve to obtain the value of the integral

Evaluate the expression: \(\int_{1}^{0}\left(2 x-x^{3}\right) d x= \frac{3}{4}\)

Key concepts

Properties of Integrals

Integrals come with specific properties that allow us to simplify and evaluate them more conveniently. Understanding these properties enhances problem-solving skills in integral calculus. One essential property is linearity, which allows for splitting apart integrals over sums and factors. This is expressed as:

• \[ \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \]
• \[ \int c \cdot f(x) \, dx = c \cdot \int f(x) \, dx \]

The first property shows that we can integrate each function separately and add the results. The second property introduces constants, highlighting how they can be factored out of the integral for simplification. Additionally, there is a property relating to limits: the reversal of limits, which changes the sign of the integral:

• \[ \int_a^b f(x) \, dx = -\int_b^a f(x) \, dx \]

Knowing these properties helps in evaluating and simplifying integrals considerably.

Integral Evaluation

Integral evaluation refers to the process of calculating the definite integral of a function. A definite integral represents the signed area under the curve of the function and has both an upper and lower limit. When dealing with integrals, it's important to identify the function's behavior within the given limits. Substituting known values or utilizing known integrals can ease this process immensely.
First, recognize the integrand, which is the function inside the integral.
In our exercise, we start with known integral values and use different properties to simplify and directly compute these new integrals.
By doing so, not only do we grasp a better understanding of how integrals operate but we also see practical applications of properties, making evaluation faster and more intuitive.

Constant Factor Rule

The constant factor rule is an important tool in calculus, especially when simplifying integrals. It states that a constant can be factored out of the integral, making the integration process much easier. For example, if a function has a constant multiplied by a variable inside an integral, the constant can be pulled outside:

• \[ \int a \cdot f(x) \, dx = a \int f(x) \, dx \]

In practical terms, this rule is particularly useful when dealing with functions that have the same form but differ in terms of constants. In our example problem, we see how the constant factor rule is used to first break down and then simplify the integration of more intricate functions by factoring out a 2, which is multiplied with the function.
Grasping this concept empowers calculus students to handle similar problems with ease, ensuring calculations remain straightforward and manageable.

Reversing Limits Property

The reversing limits property is another useful technique in integral calculus. It stems from the relationship between the direction of integration and the integral's sign. When the limits of integration are swapped, the integral's value changes sign. This can be expressed as:

• \[ \int_a^b f(x) \, dx = -\int_b^a f(x) \, dx \]

This property is very handy when dealing with integrals where the limits are perhaps reversed but the integrand remains the same.
In the given exercise, this property was particularly critical in the second part of the problem, where the direction of integration was reversed for simplification. Understanding this concept allows students to adjust integrals appropriately and can be quite insightful when approaching complex problems, revealing yet another layer of depth in the study of calculus.