EXERCISE 9.2

EXERCISE 9.2 1. $\left\{\frac{1}{n}\right\}_{n=1}^{+\infty}$ $$ \begin{aligned} & a_{n}=\frac{1}{n}, \quad a_{n+1}=\frac{1}{n+1} \\ \\ & a_{n+1}-a_{n}=\frac{1}{n+1}-\frac{1}{n} \\ \\ & \quad \quad \quad \quad =\frac{n-(n+1)}{n(n+1)} \\ \\ &\quad \quad \quad \quad =\frac{n-n-1}{n(n+1)}=-\frac{1}{n(n+1)}0 \text { For } n \geqslant 1 \\ \\ & \Rightarrow \quad a_{n+1}>a_{n} \text { For } n \geqslant 1 \end{aligned} $$ Thus $\left\{a_{n}\right\}$ is strictly … Read more

EXERCISE 9.1

EXERCISE 9.1 1. (a) $$ \begin{aligned} & 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots \\ \\ & a_{1}=1, \quad a_{2}=\frac{1}{3}, \quad a_{3}=\frac{1}{3^{2}}, a_{4}=\frac{1}{3^{3}}, \ldots \\ \\ & a_{1}=1, \quad a_{2}=\frac{1}{3^{2-1}}, \quad a_{3}=\frac{1}{3^{3-1}}, a_{4}=\frac{1}{3^{4-1}}, \ldots \\ \\ & \ldots \quad a_{n}=\frac{1}{3^{n-1}}, \ldots \end{aligned} $$ (b) $$ \begin{aligned} & 1, \quad-\frac{1}{3}, \frac{1}{9},-\frac{1}{27}, \cdots \\ \\ & a_{1}=(-1)^{1-1}, \quad a_{2}=(-1)^{2-1} … Read more

EXERCISE 7.8

3. $$ \begin{aligned} \int_{0}^{+\infty} e^{-2 x} d x & =\lim _{b \rightarrow+\infty} \int_{0}^{b} e^{-2 x} d x \\ \\ & =\lim_{b \rightarrow+\infty}\left[\frac{e^{-2 x}}{-2}\right]_{0}^{b} \\ \\ & =\lim _{b \rightarrow+\infty}\left[-\frac{e^{-2 b}}{2}+\frac{e^{0}}{2}\right] \\ \\ & =0+\frac{1}{2}=\frac{1}{2} \end{aligned} $$ 4. $$ \begin{aligned} \int_{-1}^{+\infty} \frac{x}{1+x^{2}} d x & =\lim _{b \rightarrow+\infty} \int_{-1}^{b} \frac{x}{1+x^{2}} d x \\ \\ & =\lim … Read more

EXERCISE 7.5

EXERCISE 7.5 9. $$ \begin{aligned} & \int \frac{d x}{x^{2}-3 x-4} \\ & \frac{1}{x^{2}-3 x-4}=\frac{1}{x^{2}-4 x+x-4}=\frac{1}{x(x-4)+1(x-4)} \\ & =\frac{1}{(x-4)(x+1)} \end{aligned} $$ Partial fraction decomposition $$ \frac{1}{(x-4)(x+1)}=\frac{A}{x-4}+\frac{B}{x+1} \quad \text { (1) } $$ Multiplying by $(x-4)(x+1)$ $$ 1=A(x+1)+B(x-4) \quad \text { (2) } $$ Put $x+1=0 \Rightarrow x=-1$ into (2) $$ \begin{aligned} & 1=A(-1+1)+B(-1-4) \\ & 1=-5 … Read more

Calculus Solutions EX #4.2

[embeddoc url=”https://genesismath.com/wp-content/uploads/2023/11/EXERCISE-4.2.pdf” viewer=”google”] EXERCISE 4.2 3. (a) $$f(x)=3 x^{2}-6 x+1 \quad \quad (1)$$ First derivative test Differentiating (1) w.r.t $x$. $$ f^{\prime}(x)=6 x-6 \quad \quad (2) $$ Given that $x_{0}=1.$ Sign analysis of $f^{\prime}(x)$ $$ \begin{center} \begin{tabular}{|c|c|c|c|}\hline\text {Interval} & $\begin{array}{c}\text { test point } \\ c\end{array}$ & $\begin{array}{l}f^{\prime }(c) \end{array}$ & Sign of $f^{\prime}(x)$ \\\hline$x<1$ … Read more

EXERCISE 7.4

  EXERCISE 7.4 $1-26$ Evaluate the integral. (1) $$ \int \sqrt{4-x^{2}} d x$$ $$ =\int \sqrt{2^{2}-x^{2}} —(1)d x $$ Let $$ x=2 \sin \theta—(2) $$ $$ d x=2 \cos \theta d \theta $$ then (1) becomes $$ \begin{aligned} & \int \sqrt{4-4 \sin ^{2} \theta} 2 \cos \theta d \theta \\ \\ = & 2 \int … Read more

EXERCISE 7.3

Calculus exercise 7.3

  Exercise 7.3 $1-52$ Evaluate the integral. (1) $$ \int \cos ^{3} x \sin x d x —(1)$$ $$ \begin{aligned} \text { Let } \quad u & =\cos x –(2)\\ \\ d u & =-\sin x d x \\ \\ -d u & =\sin x d x \end{aligned} $$ then (1) becomes $$ -\int u^{3} … Read more

EXERCISE 7.2

Calculus exercise 7.2

Integration by parts technique is used to solve problems in a simple way with complete steps. $1-38$ Evaluate the integral. $$\int x e^{-2 x} d x$$ Integrating by parts $$ \int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x $$ Here $u=x, v=e^{-2 x}$ $$ \begin{aligned} \therefore … Read more

EXERCISE 7.1

Calculus exercise 7.1

  1-30 Evaluate the integrals by making appropriate u-substitution. $$\quad \int(4-2 x)^{3} d x ——–(1)$$ Let $$ u=4-2 x——-(2) $$ $$ du=-2dx \Rightarrow \frac{-du}{2}=dx$$ then (1) be comes $$ \int u^{3}\left(-\frac{1}{2} d u\right)=-\frac{1}{2} \int u^{3} d u $$ Integrating w.r.t $u$, we get $$ =\frac{-1}{2}\left[\frac{u^{4}}{4}\right]+c=-\frac{1}{8} u^{4}+c—–(3) $$ $$ \text {From (2) substituting } u=4-2 x … Read more