Science:Math Exam Resources/Courses/MATH101/April 2015/Question 01 (a)

MATH101 April 2015

  •  Q1 (a)  •  Q1 (b) (i)  •  Q1 (b) (ii)  •  Q1 (b) (iii)  •  Q1 (c)  •  Q1 (d)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c) (i)  •  Q2 (c) (ii)  •  Q2 (c) (iii)  •  Q2 (c) (iv)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  Q10 (a)  •  Q10 (b)  •  Q11 (a)  •  Q11 (b)  •  

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[hide] Question 01 (a)

Only one of the following statements is always true; determine which one is true. (Assume f(x) and g(x) are continuous functions.) A. $\int _{-3}^{-2}f(x)dx=-\int _{3}^{2}f(x)dx$. B. $\int _{0}^{1}f(x)g(x)dx=\int _{0}^{1}f(x)dx\cdot \int _{0}^{1}g(x)dx.$ C. If $\int _{1}^{\infty }f(x)dx$ converges and $g(x)\geq f(x)\geq 0$ for all $x$, then $\int _{1}^{\infty }g(x)dx$ converges. D. If $\sum _{n=1}^{\infty }(-1)^{n+1}b_{n}$ converges, then $\sum _{n=1}^{\infty }b_{n}$ also converges. E, When $f(x)$ is positive and concave up, any Trapezoid Rule approximation for $\int _{a}^{b}f(x)dx$ will be an upper estimate for $\int _{a}^{b}f(x)dx$,

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[show] Hint
Plug some functions into $f(x)$ and $g(x)$ randomly and guess whether the statement is true or not.

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[show] Solution

A. $\int _{-3}^{-2}f(x)dx=-\int _{3}^{2}f(x)dx$. False. Try $f(x)=2x$. Then, $\int _{-3}^{-2}f(x)dx=x^{2}{\bigg |}_{-3}^{-2}=4-9=-5$. On the other hand, $-\int _{3}^{2}f(x)dx=\int _{2}^{3}f(x)dx=x^{2}{\bigg |}_{2}^{3}=9-4=5.$. Indeed, this would be true if $f$ is an even function. B. $\int _{0}^{1}f(x)g(x)dx=\int _{0}^{1}f(x)dx\cdot \int _{0}^{1}g(x)dx.$ False. Try $f(x)=g(x)=x$. Then, $\int _{0}^{1}f(x)g(x)dx=\int _{0}^{1}x^{2}dx={\frac {1}{3}}$. However, $\int _{0}^{1}f(x)dx\cdot \int _{0}^{1}g(x)dx=\left(\int _{0}^{1}xdx\right)^{2}={\frac {1}{4}}$. C. If $\int _{1}^{\infty }f(x)dx$ converges and $g(x)\geq f(x)\geq 0$ for all $x$, then $\int _{1}^{\infty }g(x)dx$ converges. False. Consider $f(x)=e^{-x}$ and $g(x)=1$. Apparently, $0\leq e^{-x}\leq 1$ and $\int _{1}^{\infty }e^{-x}dx=-e^{-x}{\bigg |}_{1}^{\infty }=e^{-1}<\infty$. However, $\int _{1}^{\infty }1dx=\infty$. i.e, it doesn't converge. This would be true if the order of $f(x)$ and $g(x)$ is changed in the inequality: $f(x)\geq g(x)\geq 0$. D. If $\sum _{n=1}^{\infty }(-1)^{n+1}b_{n}$ converges, then $\sum _{n=1}^{\infty }b_{n}$ also converges. False. If $b_{n}={\frac {1}{n}}$, then by the alternating convergence test, we have $\sum _{n=1}^{\infty }(-1)^{n+1}b_{n}<+\infty$. On the other hand, p-series test implies that $\sum _{n=1}^{\infty }b_{n}$ diverges. E, When $f(x)$ is positive and concave up, any Trapezoid Rule approximation for $\int _{a}^{b}f(x)dx$ will be an upper estimate for $\int _{a}^{b}f(x)dx$, True! Since $f(x)$ is positive and concave up, it satisfies $f(\theta x+(1-\theta )y)\leq \theta f(x)+(1-\theta )f(y)$ for $0<\theta <1$ and any $x<y$ on the real line. This means that the graph of $f$ on any interval $[x,y]$ never lies above of the line passing through $(x,f(x))$ and $(y,f(y))$. From this observation and the definition of Trapezoid Rule approximation, we can easily see the statement. Answer: $\color {blue}E$.