Exam Questions **Question 1:** For what values of x does the series co | Mathematics I 2074

Mathematics I207410 marksNumerical: series convergence, double integralConvergence tests and power series

Exam Questions

Question 1: For what values of x does the series converge? $\sum_{n=1}^{\infty} \frac{(x-3)^n}{x}$

Question 2: Calculate $\iint_{R} f(x, y) dA$ , for $f(x,y) = 100 - 6x^2y$ , and $R: 0 \leq x \leq 2, -1 \leq y \leq 1$ . [5+5]

Exam Questions

Question 1: For what values of x does the series converge? $$\sum_{n=1}^{\infty} \frac{(x-3)^n}{x}$$

Question 2: Calculate $\iint_{R} f(x, y) dA$, for $f(x,y) = 100 - 6x^2y$, and $R: 0 \leq x \leq 2, -1 \leq y \leq 1$. [5+5]

(a) Convergence of $\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ - Series: $\displaystyle\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ Since $x$ is independent of the summation index $n$ , factor out $\frac{1}{x}$ : $\sum{n=1}^{\infty} \frac{(x-3)^n}{x} = \frac{1}{x}\sum{n=1}^{\infty} (x-3)^n$ This requires

(a) Convergence of $\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ - Series: $\displaystyle\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ Since $x$ is independent of the summation index $n$, factor out $\frac{1}{x}$: $$\sum{n=1}^{\infty} \frac{(x-3)^n}{x} = \frac{1}{x}\sum{n=1}^{\infty} (x-3)^n$$ This requires

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Exam Questions **Question 1:** For what values of x does the series converge? **Question 2:** Calculate iintR f(x, y) dA, for f(x,y) = 100 - 6x2y, and R: 0 leq x leq 2, -1 leq y leq 1. [5+5]

Exam Questions Question 1: For what values of x does the series converge? Question 2: Calculate iintR f(x, y) dA, for f(x,y) = 100 - 6x2y, and R: 0 leq x leq 2, -1 leq y leq 1. [5+5]