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Evaluate the definite integrals. Whenever possible, use the Fundamental Theorem of Calculus, perhaps after a substitution. Otherwise, use numerical methods. $$\int_{-1}^{2} \sqrt{x+2} d x$$

Short Answer

Expert verified
The integral evaluates to \( \frac{14}{3} \).

Step by step solution

01

Understanding the Integral

The given integral is \( \int_{-1}^{2} \sqrt{x+2} \, dx \). We need to find the area under the curve of \( f(x) = \sqrt{x+2} \) from \( x = -1 \) to \( x = 2 \).
02

Identify Method of Solution

Since the integral involves a simple algebraic function, we can use the Fundamental Theorem of Calculus after finding an antiderivative for \( \sqrt{x+2} \).
03

Finding the Antiderivative

To integrate \( \sqrt{x+2} \), recognize it as \((x+2)^{1/2}\). The antiderivative is found by increasing the exponent by 1 to get \( 3/2 \) and then dividing by the new exponent. The antiderivative is \( \frac{2}{3} (x+2)^{3/2} + C \).
04

Apply the Fundamental Theorem of Calculus

The theorem states that if \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Plugging our limits into the antiderivative: \( F(2) = \frac{2}{3} (2+2)^{3/2} \) and \( F(-1) = \frac{2}{3} (-1+2)^{3/2} \).
05

Calculate F(2)

Evaluate \( F(2) = \frac{2}{3} (4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3} \).
06

Calculate F(-1)

Evaluate \( F(-1) = \frac{2}{3} (1)^{3/2} = \frac{2}{3} \times 1 = \frac{2}{3} \).
07

Compute the Definite Integral

Substitute these values back into the Fundamental Theorem expression: \( \int_{-1}^{2} \sqrt{x+2} \, dx = \frac{16}{3} - \frac{2}{3} = \frac{14}{3} \).
08

Conclusion

Thus, the value of the definite integral \( \int_{-1}^{2} \sqrt{x+2} \, dx \) is \( \frac{14}{3} \).

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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 cornerstone in the field of calculus. It's essential for connecting differentiation and integration, two of the main operations in calculus. The theorem is made up of two parts.
  • The first part states that if a function is integrable over an interval, then the integral defines a function whose derivative is the original function. This means integration can "reverse" differentiation.
  • The second part, which is more frequently used, tells us how we can compute a definite integral if we know an antiderivative of the function. If \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \).
This powerful theorem allows us to evaluate definite integrals without having to calculate the entire area under a curve by traditional means. By identifying an antiderivative, we can bypass complex calculations and still find accurate areas efficiently.
Antiderivative
An antiderivative, often interchangeably called an indefinite integral, is a function whose derivative returns the original function. In essence, finding an antiderivative is the reverse of finding a derivative.
To capture this concept, consider derivatives that measure change, while antiderivatives retrieve the original function from this measure.
When evaluating definite integrals, we need the antiderivative to compute the result. This measurement is often expressed as
  • If \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
  • The general form includes a constant \( C \), making it \( F(x) + C \) since differentiation removes constants.
  • The antiderivative of \( (x+2)^{1/2} \) involves increasing the exponent by 1 and dividing by this new exponent.
Thus, from the exercise, the antiderivative of \( \sqrt{x+2} \), which is \((x+2)^{1/2}\), is \( \frac{2}{3} (x+2)^{3/2} + C \). This step is crucial as it enables us to apply the Fundamental Theorem of Calculus later on.
Integration by Substitution
Integration by Substitution is a technique used to simplify the integration process. It's akin to the reverse of the chain rule used in differentiation. If an integral seems complicated at first glance, substitution can break it down into a simpler form.
The basic idea is to change the variable and the differential in a way that the integral becomes more manageable.
Unfortunately, not all integrals require this method. In the original exercise, a direct approach was possible without substitution, demonstrating the versatility of integration techniques.
  • Start by identifying the substitution that relates complex parts of the integral. This usually means setting \( u = g(x) \), where \( g(x) \) forms a part of the original integral.
  • Next, compute \( du \) by differentiating \( u \).
  • Rewrite the integral in terms of \( u \) and \( du \) for ease of integration.
  • Finally, once integrated, revert back to the original variable \( x \).
This method can simplify the evaluation of integrals that initially appear daunting, though in some exercises like the definite integrals one provided here, direct application of calculus principles suffices.

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Most popular questions from this chapter

(a) Between 2005 and \(2015,\) ACME Widgets sold widgets at a continuous rate of \(R=R_{0} e^{0.125 t}\) widgets per year, where \(t\) is time in years since January 1 2005\. Suppose they were selling widgets at a rate of 1000 per year on January \(1,2005 .\) How many widgets did they sell between 2005 and \(2015 ?\) How many did they sell if the rate on January 1,2005 was 1,000,000 widgets per year? (b) In the first case ( 1000 widgets per year on January 1,2005)\(,\) how long did it take for half the widgets in the ten-year period to be sold? In the second case \((1.000 .000 \text { widgets per year on January } 1,2005)\) when had half the widgets in the ten-year period been sold? (c) In \(2015,\) ACME advertised that half the widgets it had sold in the previous ten years were still in use. Based on your answer to part (b), how long must a widget last in order to justify this claim?

Explain what is wrong with the statement. The table of integrals cannot be used to find \(\int \frac{d t}{7-t^{2}}\)

Over the past fifty years the carbon dioxide level in the atmosphere has increased. Carbon dioxide is believed to drive temperature, so predictions of future carbon dioxide levels are important. If \(C(t)\) is carbon dioxide level in parts per million (ppm) and \(t\) is time in years since \(1950,\) three possible models are: \(^{1}\) I \(C^{\prime}(t)=1.3\) II \(C^{\prime}(t)=0.5+0.03 t\) III \(C^{\prime}(t)=0.5 e^{0.02 z}\) (a) Given that the carbon dioxide level was 311 ppm in \(1950,\) find \(C(t)\) for each model. (b) Find the carbon dioxide level in 2020 predicted by each model.

Explain what is wrong with the statement.To integrate $$\int \frac{1}{(x-1)^{2}(x-2)} d x$$,using a partial fraction decomposition, let $$\frac{1}{(x-1)^{2}(x-2)}=\frac{A}{(x-1)^{2}}+\frac{B}{x-2}$$.

Decide whether the statements are true or false. Give an explanation for your answer.To calculate \(\int \frac{1}{x^{3}+x^{2}} d x,\) we can split the integrand into $$\int\left(\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}\right) d x$$.

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