Problem 103 - Calculus Volume 1

Question

In the following exercises, given that \(\int_{0}^{1} x d x=\frac{1}{2}, \int_{0}^{1} x^{2} d x=\frac{1}{3}, \quad\) and \(\quad \int_{0}^{1} x^{3} d x=\frac{1}{4}\) compute the integrals. \(\int_{0}^{1}\left(7-5 x^{3}\right) d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{23}{4}\).

Step 1: Split the Integral

The first step is to split the given integral into the sum of two separate integrals. This utilizes the property of integrals allowing us to break up the integral of a sum into the sum of integrals:\[\int_0^1 (7 - 5x^3) \, dx = \int_0^1 7 \, dx - \int_0^1 5x^3 \, dx.\]

Step 2: Integrate the Constant Term

The integral of a constant function can be computed directly. For the integral \(\int_0^1 7 \, dx\), we know that the integral of \(7\) over the interval \([0, 1]\) is:\[\int_0^1 7 \, dx = 7 \int_0^1 1 \, dx = 7x \bigg|_0^1 = 7 \times (1 - 0) = 7.\]

Step 3: Apply the Given Integral

The second part of our split problem is \(\int_0^1 5x^3 \, dx\). We can use the given value \(\int_0^1 x^3 \, dx = \frac{1}{4}\) to solve it. This becomes:\[\int_0^1 5x^3 \, dx = 5 \cdot \int_0^1 x^3 \, dx = 5 \cdot \frac{1}{4} = \frac{5}{4}.\]

Step 4: Combine the Results

Finally, we combine the results from the two integrals. The first integral gave us \(7\) and the second integral gave us \(\frac{5}{4}\). Thus:\[\int_0^1 (7 - 5x^3) \, dx = 7 - \frac{5}{4}.\]

Step 5: Simplify the Expression

Simplify \(7 - \frac{5}{4}\) by converting \(7\) into a fraction with a common denominator:\[7 = \frac{28}{4},\]so,\[\frac{28}{4} - \frac{5}{4} = \frac{28 - 5}{4} = \frac{23}{4}.\]

Key Concepts

Integral Properties

When dealing with definite integrals, some properties can greatly simplify the evaluation of an expression. First and foremost, the integral of a sum can be separated into individual integrals. This property is powerful because it allows us to handle complicated expressions by breaking them down into simpler parts. For example,

• \( \int (a + b) \, dx = \int a \, dx + \int b \, dx \)

Applying these properties, as illustrated in the original step by step solution, allowed us to break down \( \int_0^1 (7 - 5x^3) \, dx \) into two separate integrals:

• \( \int_0^1 7 \, dx \)
• \( - \int_0^1 5x^3 \, dx \)

This decomposition made it easy to evaluate each part using known values or easier computation steps.

Integral of a Constant

Computing the integral of a constant function is straightforward. When integrating a constant \( c \) over an interval \([a, b]\), the result is simply that constant multiplied by the width of the interval. Mathematically, it is depicted as follows:

• \( \int_a^b c \, dx = c \cdot (b - a) \)

In the given exercise, for \( \int_0^1 7 \, dx \), the constant 7 is integrated over the interval \([0, 1]\). Using the property, we get:

• \( 7 \cdot (1 - 0) = 7 \)

This simplification technique confirms how simple it is to evaluate integrals of constant values. It reduces the problem to basic multiplication, which is quick to both write and understand.

Combining Integrals

After splitting an integral, the last steps involve calculating individual integrals and then recombining the results. Bringing together solutions from previously split integrals helps in achieving the final answer. In the context of our problem, after finding:

• \( \int_0^1 7 \, dx = 7 \)
• \( -\int_0^1 5x^3 \, dx = -\frac{5}{4} \)

The next step is to combine these results back together, thus:

• \( 7 - \frac{5}{4} \)

This step demonstrates combining answers of evaluated integrals into a single expression, helping us reach the solution effectively and systematically.

Fraction Simplification

Combining results from different computed integrals often ends with expressions needing simplification, such as subtracting fractions. Simplifying fractions follows algebraic rules that often require creating common denominators.When converting integer results into fractions to perform operations such as addition or subtraction, the integer can be rewritten with an equivalent fraction. In this context, \(7\) was converted to \(\frac{28}{4}\), thus sharing a common denominator with \(\frac{5}{4}\). Subtraction was then straightforward:

• \( \frac{28}{4} - \frac{5}{4} = \frac{23}{4} \)

These processes exemplify how fractional operations can be managed by ensuring denominators match, making fractions easy to add or subtract. This final step is crucial for expressing results neatly and comprehensively.