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Chapter 5: Problem 5

Evaluate the integrals in Exercises \(1-28\). $$\int_{1}^{4}\left(3 x^{2}-\frac{x^{3}}{4}\right) d x$$

Short Answer

Expert verified
The integral evaluates to 47.0625.

Step by step solution

01

Identify the integral to solve

The given integral is \( \int_{1}^{4}\left(3 x^{2}-\frac{x^{3}}{4}\right) d x \). This is an integral of a polynomial function over the interval from \( x=1 \) to \( x=4 \).
02

Apply the Power Rule

To find the antiderivatives of each term, use the power rule: for \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} \).
03

Antidifferentiate Each Term

For \( 3x^2 \), the antiderivative is \( \frac{3x^{3}}{3} = x^3 \). \[\text{For } -\frac{x^3}{4}, \text{ the antiderivative is } -\frac{x^4}{16}. \]
04

Write the Antiderivative Function

Combine the results from Step 3 to write the antiderivative: \[ F(x) = x^3 - \frac{x^4}{16}. \]
05

Evaluate the Definite Integral

Use the Fundamental Theorem of Calculus: evaluate \( F(x) \) at the upper and lower bounds, and subtract: \[ F(4) = 4^3 - \frac{4^4}{16} \]\[ F(1) = 1^3 - \frac{1^4}{16}. \]
06

Calculate the Values

Calculate \( F(4) \): \[ 4^3 - \frac{4^4}{16} = 64 - 16 = 48. \]Calculate \( F(1) \): \[ 1^3 - \frac{1}{16} = 1 - 0.0625 = 0.9375. \]
07

Find the Difference

Subtract to find the value of the definite integral: \[ F(4) - F(1) = 48 - 0.9375 = 47.0625. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a vital bridge between differentiation and integration. It comes in two parts. First, it tells us that if we integrate a function over an interval, we can find the total change by evaluating its antiderivative at the endpoints of the interval.
Simply put, if you have a function's antiderivative, you can calculate the area under the curve between two points by subtracting the antiderivative's values at these points.
This is expressed as:
  • Find an antiderivative of the function.
  • Evaluate this antiderivative at the upper limit of the interval.
  • Evaluate the antiderivative at the lower limit of the interval.
  • Subtract the second result from the first.
This process allows you to compute a definite integral without manually summing infinitely small pieces. It's like finding the distance between two points by looking at a map rather than walking every step.
Power Rule for Integration
The power rule for integration is a straightforward technique used to find the antiderivatives of polynomial terms. It states that to integrate a term like \( x^n \), you increase the exponent by one and then divide by the new exponent.
This transforms into:
  • If \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} \).
  • Always remember to adjust any coefficients, such as in \( 3x^2 \), the process involves \( \frac{3x^{3}}{3} \).
  • Don't forget to apply the same rule to any negative or fractional powers.
Using this rule efficiently, you can solve polynomial integrals by applying it term by term. It's a handy tool because much of calculus involves polynomial expressions.
Polynomial Integration
Polynomial integration is all about applying the fundamental tools of calculus to polynomial expressions. A polynomial is simply a sum of terms like \( ax^n \). During integration:
  • Each term of the polynomial is treated separately.
  • Use the power rule for each term to find their respective antiderivatives.
  • Combine these antiderivatives to form a single function.
By breaking down the polynomial into its simplest terms, you can tackle each part independently.
This simplification makes the daunting task of integration more approachable. Let鈥檚 take \(3x^2 - \frac{x^3}{4}\) as an example:
  • First, find \(3x^2\)'s antiderivative: \(x^3\).
  • Second, integrate \(-\frac{x^3}{4}\): \(-\frac{x^4}{16}\).
  • Combine these results to get the total antiderivative of the polynomial.
This method is efficient and shows the power of calculus in dealing with various equations effortlessly.

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Most popular questions from this chapter

In Exercises \(47-50,\) find the total area between the region and the \(x\) -axis. $$y=-x^{2}-2 x, \quad-3 \leq x \leq 2$$

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