Question

i) A particle moves along a line with velocity function v(t)= (t² - t), where v is measured in meters per second. Find (a) the displacement and (b) the distance traveled by

the particle during the time interval [ 0, 5]

ii) Use the properties of integrals to verify the inequality.

Integral of $\intop (sinx/x) dx$ Upper limit ($\pi$ /2) and Lower limit ($\pi$ /4) <= ( √2 / 2) .

iii).Evaluate the intregal, if it exists.

1. $\intop (1+ tan t)$ ³ sec²t dt ( Upper limit ($\pi$ /4) and Lower limit (0)
2. $\intop\begin{vmatrix} √x - 1 \\ \end{vmatrix}$ dx (Upper limit (4) and Lower limit (0)
3. $\intop$tan x ln(cos x) dx

Answer 1

i)

a)

$$s(t)=\int v(t)dt=\int(t^2-t)dt=\dfrac{t^3}{3}-\dfrac{t^2}{2}+s(0),m$$

b)

$$s(5)=\dfrac{(5)^3}{3}-\dfrac{(5)^2}{2}+s(0)=\dfrac{175}{6}+s(0) (m)$$

$$distance=s(5)-s(0)=\dfrac{175}{6}+s(0) -s(0)$$

$$=\dfrac{175}{6}(m)$$

ii)

$$\sin x=\displaystyle\sum_{i=0}^{\infin}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$

$$\dfrac{\sin x}{x}=\displaystyle\sum_{i=0}^{\infin}\dfrac{(-1)^{n}x^{2n}}{(2n+1)!}$$

$$=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+...$$

Integrate term by term

$$\int\dfrac{\sin x}{x}dx=\int\displaystyle\sum_{i=0}^{\infin}\dfrac{(-1)^{n}x^{2n}}{(2n+1)!}dx$$

$$=\displaystyle\sum_{i=0}^{\infin}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!(2n+1)}+C$$

$$=C+x-\dfrac{x^3}{3!(3)}+\dfrac{x^5}{5!(5)}-\dfrac{x^7}{7!(7)}+...$$

Use the second fundamental theorem of calculus to evaluate

$$\displaystyle\int_{\pi/4}^{\pi/2}\dfrac{\sin x}{x}dx=\displaystyle\int_{\pi/4}^{\pi/2}\displaystyle\sum_{i=0}^{\infin}\dfrac{(-1)^{n}x^{2n}}{(2n+1)!}dx$$

$$=\bigg[x-\dfrac{x^3}{3!(3)}+\dfrac{x^5}{5!(5)}-\dfrac{x^7}{7!(7)}+...\bigg]\begin{matrix} \pi/2 \\ \pi/4 \end{matrix}$$

$$\leq\dfrac{\pi}{2}-\dfrac{(\dfrac{\pi}{2})^3}{18}+\dfrac{(\dfrac{\pi}{2})^5}{600}-(\dfrac{\pi}{4}-\dfrac{(\dfrac{\pi}{4})^3}{18}+\dfrac{(\dfrac{\pi}{4})^5}{600})$$

$$\leq0.6125\leq\dfrac{\sqrt{2}}{2}$$

Therefore

$$\displaystyle\int_{\pi/4}^{\pi/2}\dfrac{\sin x}{x}dx\leq\dfrac{\sqrt{2}}{2}$$

iii)

1.

$$\int(1+\tan t)^3\sec^2tdt$$

$$u=1+\tan t, du=\sec^2t dt$$

$$\int(1+\tan t)^3\sec^2tdt=\int u^3du=\dfrac{u^4}{4}+C$$

$$=\dfrac{(1+\tan t)^4}{4}+C$$

2.

$$\displaystyle\int_{0}^{4}|\sqrt{x}-1|dx=-\displaystyle\int_{0}^{1}(\sqrt{x}-1)dx+\displaystyle\int_{1}^{4}(\sqrt{x}-1)dx$$

$$=-[\dfrac{2x^{3/2}}{3}-x]\begin{matrix} 1 \\ 0 \end{matrix}+[\dfrac{2x^{3/2}}{3}-x]\begin{matrix} 4 \\ 1 \end{matrix}$$

$$=-\dfrac{2}{3}+1+0+\dfrac{16}{3}-4-\dfrac{2}{3}+1=2$$

3.

$$\int \tan x\ln(\cos x)dx$$

$$u=\ln(\cos x), du=\dfrac{1}{\cos x}(-\sin x)dx=-\tan xdx$$

$$\int \tan x\ln(\cos x)dx=-\int udu=-\dfrac{u^2}{2}+C$$

$$=-\dfrac{\ln^2(\cos x)}{2}+C$$