/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Compute \(\int\left(1+t^{4}\righ... [FREE SOLUTION] | 91Ó°ÊÓ

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Compute \(\int\left(1+t^{4}\right)^{3} d t\).

Short Answer

Expert verified
The integral is \( t + \frac{3}{5}t^5 + \frac{1}{3}t^9 + \frac{1}{13}t^{13} + C \).

Step by step solution

01

Identify the integral form

The integral given is \( \int \left(1 + t^4\right)^3 \, dt \). This is an integral of a polynomial function raised to a power, which suggests using expansion followed by term-by-term integration.
02

Expand the polynomial

Expand the expression \( \left(1 + t^4\right)^3 \) using the binomial theorem. This gives \[\left(1 + t^4\right)^3 = \binom{3}{0} \cdot 1^3 + \binom{3}{1} \cdot 1^2 \cdot t^4 + \binom{3}{2} \cdot 1 \cdot (t^4)^2 + \binom{3}{3} \cdot (t^4)^3\]This simplifies to \[1 + 3t^4 + 3t^8 + t^{12}\].
03

Integrate each term separately

Integrate the polynomial term by term: \[\int (1 + 3t^4 + 3t^8 + t^{12}) \, dt = \int 1 \, dt + \int 3t^4 \, dt + \int 3t^8 \, dt + \int t^{12} \, dt\]Compute each integral:1. \( \int 1 \, dt = t \)2. \( \int 3t^4 \, dt = 3 \cdot \frac{t^5}{5} = \frac{3}{5}t^5 \)3. \( \int 3t^8 \, dt = 3 \cdot \frac{t^9}{9} = \frac{1}{3}t^9 \)4. \( \int t^{12} \, dt = \frac{t^{13}}{13} \).
04

Combine the integrated terms

Add the results of the separate integrals together, including the constant of integration \( C \):\[t + \frac{3}{5}t^5 + \frac{1}{3}t^9 + \frac{1}{13}t^{13} + C\]This is the solution to the integral \( \int \left(1 + t^4\right)^3 \, dt \).

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Key Concepts

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

Polynomial Integration
Polynomial integration is a fundamental concept in calculus that involves finding the antiderivative of polynomial functions. In simpler terms, it's about reversing the process of differentiation for polynomial expressions. The main goal here is to find a function whose derivative would give us the original polynomial we started with. When dealing with polynomials, each term is integrated separately, and the integration of powers of a variable is done by applying the power rule in reverse. This power rule for integration states that the integral of \( t^n \) is \( \frac{t^{n+1}}{n+1} \), given that \( n eq -1 \).
For instance, integrating a polynomial like \( 3t^8 \) involves adding one to the exponent (making it 9) and then dividing by this new exponent (resulting in \( \frac{1}{3}t^9 \) after the multiplication by the coefficient 3 is incorporated). Each term is treated in this way, and then all antiderivatives are added together, with a constant of integration \( C \) added to represent the family of antiderivatives that can differ by a constant.
Binomial Theorem
The Binomial Theorem provides a way to expand expressions that are raised to a power, such as \( (1 + t^4)^3 \). This theorem tells us how to expand a binomial expression and is particularly useful in polynomial integration when dealing with terms raised to a power.
The binomial theorem expresses \((a + b)^n\) as a sum involving terms of the form \( \binom{n}{k} a^{n-k} b^k \), where \( \binom{n}{k} \) are binomial coefficients. These coefficients can be found using Pascal's triangle or calculated directly as \( \frac{n!}{k!(n-k)!} \).
  • For \((1 + t^4)^3\), the binomial expansion results in: \( 1 + 3t^4 + 3t^8 + t^{12} \).
  • This step simplifies our polynomial expression before integration.
By expanding the expression, we can easily integrate each term separately, simplifying the entire process of finding the antiderivative.
Antiderivative Computation
Antiderivative computation involves determining the function from which a given function was derived. Integrating a function is finding its antiderivative, essentially reconstructing the original function, except here we derive a function that gives us back the polynomial upon differentiation.
Given that we have already expanded \((1 + t^4)^3\), we calculate the integral for each term in the polynomial.
  • Therefore, integrating the term \(1\) results in \(t\).
  • For \(3t^4\), the integral becomes \( \frac{3}{5}t^5 \).
  • Meanwhile, \(3t^8\) integrates to \( \frac{1}{3}t^9 \).
  • Finally, \(t^{12}\) integrates to \( \frac{1}{13}t^{13} \).
Combine these results into the complete antiderivative, and don't forget to add a constant \(C\) at the end, representing the infinite number of antiderivatives differing by a constant.
Polynomial Expansion
Polynomial expansion refers to the process of expressing a polynomial raised to a power in its expanded form. This concept is crucial when dealing with expressions like \((1 + t^4)^3\), where direct integration is not straightforward.
Using techniques like the binomial theorem allows us to write the polynomial as a sum of its expanded terms. This step involves:
  • Applying the binomial theorem, where we expand \((1 + t^4)^3\) into terms \(1 + 3t^4 + 3t^8 + t^{12}\).
  • Breaking down the complex power into manageable, integrable components.
Such expansion is critical because it converts the expression into simpler monomial forms, each of which can be integrated individually using basic polynomial integration techniques.

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Most popular questions from this chapter

Suppose the rate of glucose production in a corn plant is proportional to sunlight intensity and can be approximated by $$ R(t)=K(t+7)^{2}(t-7)^{2}=K\left(t^{4}-98 t^{2}+2401\right) \quad-7 \leq t \leq 7 $$ Time is measured so that sunrise is at -7 hours, the sun is at its zenith at 0 hours and sets at 7 hours. The quantity \(Q(x)\) of glucose produced during the period \([-7, x]\) is $$ Q(x)=\int_{-7}^{x} R(t) d t=\int_{-7}^{x} K\left(t^{4}-98 t^{2}+2401\right) d t=K \int_{-7}^{x}\left(t^{4}-98 t^{2}+2401\right) d t $$ By the Fundamental Theorem of Calculus, $$ Q^{\prime}(x)=K\left(x^{4}-98 x^{2}+2401\right) $$ a. Sketch the graph of \(R(t)\). At what time is the sun most intense? b. Show that if \(U(x)=\frac{x^{5}}{5}\) then \(U^{\prime}(x)=x^{4}\). c. Find an example of a function, \(V(x)\) such that \(V^{\prime}(x)=-98 x^{2}\). d. Find an example of a function, \(W(x)\) such that \(W^{\prime}(x)=2401\). e. Let \(G(x)=K\left(\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right)\). Show that \(G^{\prime}(x)=Q^{\prime}(x)\). f. Conclude that there is a number, \(C\), such that $$ Q(x)=G(x)+C=K\left[\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right]+C $$ a. Sketch the graph of \(R(t)\). At what time is the sun most intense? b. Show that if \(U(x)=\frac{x^{5}}{5}\) then \(U^{\prime}(x)=x^{4}\). c. Find an example of a function, \(V(x)\) such that \(V^{\prime}(x)=-98 x^{2}\). d. Find an example of a function, \(W(x)\) such that \(W^{\prime}(x)=2401\). e. Let \(G(x)=K\left(\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right)\). Show that \(G^{\prime}(x)=Q^{\prime}(x)\). f. Conclude that there is a number, \(C,\) such that $$ Q(x)=G(x)+C=K\left[\frac{x^{5}}{5}-\frac{98}{3} x^{3}+2401 x\right]+C $$ g. Why is \(Q(-7)=0\). h. Evaluate \(C\). i. Compute \(Q(7),\) the amount of glucose produced during the day.

Solve using integration by parts, \(\int u(x) v^{\prime}(x) d x=u(x) v(x)-\int v(x) u^{\prime}(x) d x\) (or \(\left.\int u d v=u v-\int v d u\right)\) a. \(\int x e^{x} d x\) b. \(\int x \ln x d x\) c. \(\int x \sin x d x\) d. \(\int x^{2} e^{x} d x\) e. \(\int x e^{2 x} d x\) f. \(\int \ln x \cdot 1 d x\) g. \(\int x \cos x d x\) h. \(\int x^{3} e^{x^{2}} d x\)

Let \(\mathrm{G}(\mathrm{x})\) be the area of the region bounded by the graphs of \(y=\frac{1}{1+t^{2}}, y=0,\) \(t=1\) and \(t=x .\) a. Compute approximate values for \(y(1.0), y(1.5), y(2.0), y(2.5),\) and \(y(3.0)\). b. Compute approximate values for \(G(1.0), G(1.5), G(2.0), G(2.5),\) and \(G(3.0)\). c. Sketch the graph of \(y=\frac{1}{1+t^{2}}\) on [0,3] . d. Sketch the graph of \(G\) on [1,3] . e. Estimate \(G^{\prime}(2)\).

Evaluate the integrals. a. \(\int_{1}^{3} t^{2} d t\) b. \(\int_{0}^{2} t^{3} d t\) c. \(\int_{0}^{2} e^{t} d t\) d. \(\int_{1}^{3} \frac{1}{t} d t\) e. \(\int_{1}^{3}(1+t)^{2} d t\) f. \(\int_{1}^{3} 5 t d t\) g. \(\int_{0}^{2} t+5 d t\) $$ \text { h. } \int_{0}^{2} e^{2 t} d t \quad \text { i. } \quad \int_{0}^{3} e^{-t} d t \quad \text { j. } \quad \int_{1}^{3}\left(1+t^{2}\right) d t $$

Equal quantities of gaseous hydrogen and iodine are mixed resulting in the reaction $$ \mathrm{H}_{2}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{HI} $$ which runs until \(I_{2}\) is exhausted \(\left(H_{2}\right.\) is also exhausted). The rate at which \(I_{2}\) disappears is \(\frac{0.2}{(t+1)^{2}}\) \(\mathrm{gm} / \mathrm{sec} .\) How much \(I_{2}\) was initially introduced into the mixture? a. Sketch the graph of the reaction rate, \(r(t)=\frac{0.2}{(t+1)^{2}}\). b. Approximately how much \(I_{2}\) combined with \(H_{2}\) during the first second? c. Approximately how much \(I_{2}\) combined with \(H_{2}\) during the second second? d. Let \(Q(x)\) be the amount of \(I_{2}\) that combines with \(H_{2}\) during time 0 to \(x\) seconds. Write an integral that is \(Q(x)\) e. What is \(Q^{\prime}(x) ?\) f. Compute \(W^{\prime}(x)\) for \(W(x)=\frac{-0.2}{1+x}\). g. Show that there is a number, \(C,\) for which \(Q(x)=W(x)+C\). h. Show that \(C=0.2\) so that \(Q(x)=0.2-\frac{0.2}{1+x}\). i. How much \(I_{2}\) combined with \(H_{2}\) during the first second? j. How much \(I_{2}\) combined with \(H_{2}\) during the first 100 seconds? k. How much \(I_{2}\) combined with \(H_{2} ?\)

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