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Chapter 5: Problem 13

use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{1}\left(2 x^{4}-3 x^{2}+5\right) d x $$

Short Answer

Expert verified
The value of the integral is \(\frac{22}{5}\).

Step by step solution

01

Identify the Antiderivative

The first step in evaluating a definite integral using the Second Fundamental Theorem of Calculus is to find the antiderivative of the integrand. The integrand is \( 2x^4 - 3x^2 + 5 \). To find the antiderivative, integrate each term separately:- The antiderivative of \(2x^4\) is \(\frac{2}{5}x^5\).- The antiderivative of \(-3x^2\) is \(-x^3\).- The antiderivative of \(5\) is \(5x\).Combining these, the antiderivative \(F(x)\) is \(\frac{2}{5}x^5 - x^3 + 5x\).
02

Apply the Fundamental Theorem of Calculus

According to the Second Fundamental Theorem of Calculus, the definite integral from \(a\) to \(b\) of \(f(x)\) is \(F(b) - F(a)\), where \(F(x)\) is the antiderivative found in the previous step. Here, \(a = 0\) and \(b = 1\).
03

Evaluate the Antiderivative at the Bounds

Compute \(F(1)\):\[F(1) = \frac{2}{5}(1)^5 - (1)^3 + 5(1) = \frac{2}{5} - 1 + 5 = \frac{2}{5} - \frac{5}{5} + \frac{25}{5} = \frac{22}{5}\]Compute \(F(0)\):\[F(0) = \frac{2}{5}(0)^5 - (0)^3 + 5(0) = 0\]
04

Subtract the Results

Subtract the evaluation at the lower bound from the evaluation at the upper bound:\[F(1) - F(0) = \frac{22}{5} - 0 = \frac{22}{5}\]

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Key Concepts

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

Second Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus is a crucial concept that connects differentiation and integration. It asserts that if you have a continuous function, such as \[f(x) = 2x^4 - 3x^2 + 5,\] its antiderivative, denoted as \(F(x)\), can be used to calculate the definite integral over an interval from \(a\) to \(b\).
  • This theorem simplifies the process by allowing you to evaluate the function's antiderivative at specific bounds.
  • Specifically, it states that the definite integral of \(f(x)\) from \(a\) to \(b\) is \(F(b) - F(a)\).
  • By finding the antiderivative and knowing your limits, you can determine the accumulated area under the curve of \(f(x)\) between \(a\) and \(b\).
This theorem is a powerful tool in calculus, particularly for solving problems involving definite integrals.
Antiderivative
Antiderivatives are at the heart of integral calculus and are essentially the reverse of derivatives. Given a function \(f(x)\), its antiderivative \(F(x)\) satisfies the property that when you differentiate \(F(x)\), you get back \(f(x)\).
  • To find the antiderivative of a polynomial function, integrate each term separately. This involves increasing the exponent of each term by one and dividing by the new exponent. For example: \[\int 2x^4 \, dx = \frac{2}{5}x^5.\]
  • Consider \(2x^4 - 3x^2 + 5\). Its antiderivative is: \[F(x) = \frac{2}{5}x^5 - x^3 + 5x.\]
Finding the antiderivative is a preparatory step to apply the Second Fundamental Theorem of Calculus. Proper integration techniques are essential here.
Calculus Problem Solving
Solving calculus problems involves a systematic approach that combines different rules and theorems of calculus.
  • Identify the function that you need to integrate over a specific interval.
  • Determine its antiderivative, understanding that each term is treated separately in polynomial functions.
  • Apply the Second Fundamental Theorem of Calculus to evaluate the definite integral by substituting the interval bounds into the antiderivative.
  • Subtract the lower bound evaluation from the upper bound evaluation.
This step-by-step method helps in breaking down complex integration problems into manageable parts, leading to effective solutions.
Integral Evaluation
Evaluating a definite integral involves calculating the net area under the curve represented by the integrand over a specific interval.
  • Start by identifying the antiderivative, an essential process to transition from the function back to an expression that can be used for evaluation.
  • For the definite integral \( \int_{0}^{1}(2x^4 - 3x^2 + 5) \, dx\), we found that the antiderivative is \( F(x) = \frac{2}{5}x^5 - x^3 + 5x \).
  • Evaluate this function at the bounds: \( F(1) = \frac{22}{5} \) and \( F(0) = 0 \).
  • Finally, subtract to find the evaluated integral: \( F(1) - F(0) = \frac{22}{5} \).
The process of integral evaluation results in a numerical value representing the area of concern, showcasing the utility and precision of calculus in solving real-world problems.

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Most popular questions from this chapter

Use symmetry to help you evaluate the given integral. $$ \int_{-100}^{100}\left(v+\sin v+v \cos v+\sin ^{3} v\right)^{5} d v $$

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Suppose that an object, moving along the \(x\) -axis, has velocity \(v=t^{2}\) meters per second at time \(t\) seconds. How far did it travel between \(t=3\) and \(t=5\) ? See Problem 61 .
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