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Chapter 7: Problem 18

Find the integrals. $$\int(t+2) \sqrt{2+3 t} d t$$

Short Answer

Expert verified
The integral is \( \frac{1}{9} \left( \frac{2}{5} (2+3t)^{5/2} + \frac{8}{3} (2+3t)^{3/2} \right) + C \).

Step by step solution

01

Identify the Integration Technique

This integral seems to be suitable for a substitution method due to the composition of functions: a polynomial inside a square root. We will use substitution to simplify the integral.
02

Choose the Substitution

Let us substitute the inner function of the square root. Let \( u = 2 + 3t \), which makes \( du = 3 \, dt \).
03

Solve for dt

From \( du = 3 \, dt \), solve for \( dt \): \[ dt = \frac{du}{3} \]
04

Express t in terms of u

Since \( u = 2 + 3t \), solve for \( t \): \[ t = \frac{u - 2}{3} \]
05

Substitute in the Integral

Substitute \( t \) and \( dt \) in the integral: \[ \int (t+2) \sqrt{u} \, dt = \int \left(\frac{u-2}{3}+2\right) \sqrt{u} \cdot \frac{du}{3} \] Simplify: \[ \int \left(\frac{u-2+6}{3}\right) \sqrt{u} \, \frac{du}{3} = \frac{1}{9} \int (u+4) \sqrt{u} \, du \]
06

Simplify the Integral

Break down the integral: \[ \int (u+4) \sqrt{u} \, du = \int u^{3/2} \, du + 4 \int u^{1/2} \, du \]
07

Integrate Each Term Separately

Perform the integration for each term:\[ \int u^{3/2} \, du = \frac{2}{5} u^{5/2} \] \[ 4 \int u^{1/2} \, du = 4 \times \frac{2}{3} u^{3/2} = \frac{8}{3} u^{3/2} \]
08

Combine the Results and Substitute Back

Combine the results of integration: \[ \frac{2}{5} u^{5/2} + \frac{8}{3} u^{3/2} \] Substitute \( u = 2 + 3t \) back into the expression and multiply by \( \frac{1}{9} \): \[ \frac{1}{9} \left( \frac{2}{5} (2+3t)^{5/2} + \frac{8}{3} (2+3t)^{3/2} \right) + C \]
09

Simplify Final Expression

Simplify and combine terms if possible, leaving the expression in its simplest form. The integration constant \( C \) is appended as usual.

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Key Concepts

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

Substitution Method
The substitution method is a valuable technique in integral calculus, often used to simplify the integration of complex functions. It works by substituting a part of the integrand (the function being integrated) with a new variable, usually simplifying the integral to a more recognizable form.

To effectively use this technique, follow these steps:
  • Identify a part of the function that can be substituted. Typically, this involves looking for composite functions (functions within functions), as seen with square roots or trigonometric functions.
  • Choose a substitution variable, say \( u \). Calculate its differential \( du \) and solve for \( dt \) or \( dx \), depending on the variable of integration.
  • Substitute these expressions back into the integral, changing the limits if necessary when dealing with a definite integral.
  • Integrate in terms of the new variable \( u \), and finally substitute back the original expression to revert the variable change.
This method simplifies the integral, especially when faced with complex inner functions that don't integrate easily using standard rules.
Integral Calculus
Integral calculus is a branch of calculus focused on the accumulation of quantities and the areas under curves. It is fundamentally the reverse operation of differentiation.

In the context of indefinite integrals, we are finding a function whose derivative is the given function. Definite integrals, on the other hand, calculate the accumulated sum over an interval, providing a numerical result rather than an algebraic expression.

The power rule is a standard tool in integral calculus for integrating polynomials. It states that for any real number \( n \),
  • \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).
  • There is an exception when encountering a fraction or a constant multiplier, which can be factored out for simplified integration.
Understanding these basic techniques in integral calculus equips you to tackle more complex integrals effectively.
Definite Integrals
Definite integrals extend the concept of integration by calculating the net area between the curve of the function and the axes over a specified interval. It involves using limits to evaluate the antiderivative at the upper and lower bounds of the interval.

To solve a definite integral, follow these expanded steps:
  • Find the indefinite integral, which is the antiderivative of the function.
  • Use the Fundamental Theorem of Calculus. This theorem bridges the relationship between differentiation and integration, allowing you to evaluate definite integrals by subtracting the value of the antiderivative at the lower limit from its value at the upper limit.
  • For a function \( f(x) \) over the interval \([a, b]\), compute \( F(b) - F(a) \), where \( F \) is the antiderivative of \( f \).
Definite integrals are used in various applications, from calculating areas to solving physics problems involving motion and accumulation of quantities, making them invaluable in real-world problem solving.

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Most popular questions from this chapter

(a) Find \(\int \sin \theta \cos \theta d \theta\). (b) You probably solved part (a) by making the substitution \(w=\sin \theta\) or \(w=\cos \theta .\) (If not, go back and do it that way.) Now find \(\int \sin \theta \cos \theta d \theta\) by making the other substitution. (c) There is yet another way of finding this integral which involves the trigonometric identities \(\sin (2 \theta)=2 \sin \theta \cos \theta\) \(\cos (2 \theta)=\cos ^{2} \theta-\sin ^{2} \theta\). Find \(\int \sin \theta \cos \theta d \theta\) using one of these identities and then the substitution \(w=2 \theta\). (d) You should now have three different expressions for the indefinite integral \(\int \sin \theta \cos \theta d \theta .\) Are they really different? Are they all correct? Explain.

The voltage, \(V\), in an electrical outlet is given as a function of time, \(t,\) by the function \(V=V_{0} \cos (120 \pi t)\) where \(V\) is in volts and \(t\) is in seconds, and \(V_{0}\) is a positive constant representing the maximum voltage. (a) What is the average value of the voltage over 1 second? (b) Engineers do not use the average voltage. They use the root mean square voltage defined by \(\bar{V}=\) \(\sqrt{\text { average of }\left(V^{2}\right) . \text { Find } \bar{V} \text { in terms of } V_{0}, \text { (Take the }}\) average over 1 second.) (c) The standard voltage in an American house is 110 volts, meaning that \(\bar{V}=110 .\) What is \(V_{0} ?\)

Find the exact area. Under \(f(x)=1 /(x+1)\) between \(x=0\) and \(x=2\).

An electric current, \(I(t),\) flowing out of a capacitor, decays according to \(I(t)=I_{0} e^{-t},\) where \(t\) is time. Find the charge, \(Q(t),\) remaining in the capacitor at time \(t\) The initial charge is \(Q_{0}\) and \(Q(t)\) is related to \(I(t)\) by $$ Q^{\prime}(t)=-I(t) $$

In Problems \(49-50\), explain what is wrong with the statement. If both \(\int_{1}^{\infty} f(x) d x\) and \(\int_{1}^{\infty} g(x) d x\) diverge, then so \(\operatorname{does} \int_{1}^{\infty} f(x) g(x) d x\).
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