lecture 15 applications of antidiff

Find the value of a, where a>0 such that the area enclosed by the curve of function f(x) and the x-axis equals 1, where $f:[0,\pi ]\to R$, $f(x)=a\sin (x)$.

The area enclosed is ${\int }_0^{\pi }a\sin (x)\cdot dx=[ax-\cos (x){]}_0^{\pi }=1$
$a(-\cos (\pi )-(-1))=1$
$2a=1$
$a=\frac{1}{2}$

Evaluate ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\sin (2x)\cdot dx$.

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\sin (2x)\cdot dx$
$=[\frac{-\cos (2x)}{2}{]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}$
$=\frac{1}{2}-\frac{1}{2}$
$=0$

Given ${\int }_0^{3}q(x)\cdot dx=5$, find ${\int }_0^6[q\left(\frac{x}{2}\right)+x]\cdot dx$.

${\int }_0^6\left(q\left(\frac{x}{2}\right)+x\right)\cdot dx$
$={\int }_0^{6}q\left(\frac{x}{2}\right)\cdot dx+{\int }_0^{6}x\cdot dx$
$=[2\cdot Q(\frac{x}{2}){]}_0^6+[\frac{x^2}{2}{]}_0^6$
$=2\cdot [Q(x){]}_0^3+[\frac{x^2}{2}{]}_0^6$
$=2\cdot (Q(3)-Q(0))+[\frac{x^2}{2}{]}_0^6$
$=10+18$
$=28$

Evaluate ${\int }_{\frac{-\pi }{3}}^{\frac{\pi }{2}}{\sec }^2(\frac{x}{2})\cdot dx$.

$=[2\tan (\frac{x}{2}){]}_{\frac{-\pi }{3}}^{\frac{\pi }{2}}$
$=2\tan (\frac{\pi }{4})-2\tan (\frac{-\pi }{6})$
$=2+\frac{2\sqrt{3}}{3}$

Find the exact area of the region bounded by $y=\frac{1}{2}x+1$, $y=-x+2$ and the x−axis.

Intersection at $\frac{1}{2}x+1=-x+2$
$x+2=-2x+4$
$3x=2$
$x=\frac{2}{3}$

x-int at $x=-2$ and $x=2$
Total area needed is ${\int }_{-2}^{\frac{2}{3}}[\frac{1}{2}x+1]\cdot dx+{\int }_{\frac{2}{3}}^2[-x+2]\cdot dx$
$=[\frac{x^2}{4}+x{]}_{-2}^{\frac{2}{3}}+[\frac{-x^2}{2}+2x{]}_{\frac{2}{3}}^2$

Ok this is too bad. Let’s be smart and just calculate the triangle. $\frac{1}{2}\cdot 4\cdot \frac{4}{3}=\frac{8}{3}$

By first differentiating $\sqrt{x^3+1}$, find an antiderivative of $\frac{x^2}{\sqrt{x^3+1}}$.

$\frac{d}{dx}[\sqrt{x^3+1}]=\frac{1}{2}(x^3+1{)}^{\frac{-1}{2}}\cdot 3x^2=\frac{3x^2}{2\sqrt{x^3+1}}$

$\frac{3}{2}\cdot \frac{x^2}{\sqrt{x^3+1}}=\frac{d}{dx}[\sqrt{x^3+1}]$

$\frac{x^2}{\sqrt{x^3+1}}=\frac{d}{dx}[\sqrt{x^3+1}]\cdot \frac{2}{3}$

Take antiderivative of both sides

Answer is $\frac{2}{3}\cdot \sqrt{x^3+1}+C$

Find the area enclosed by the graphs $y=e^x$, $y=e^{-x}$ and $x=\ln (2)$.

Bounded area is top minus bottom

${\int }_0^{\ln (2)}{e}^x-e^{-x}\cdot dx$
$=[e^x+e^{-x}{]}_0^{\ln (2)}$
$=2+\frac{1}{2}-(2)$
$=\frac{1}{2}$

Find the average value of the function $f(x)=\frac{3}{2x-5}-3$ over the interval $[0,1]$.

Average value is ${\int }_0^1\frac{3}{2x-5}-3\cdot dx$
$={\int }_0^{1}3\cdot (2x-5{)}^{-1}-3\cdot dx$
$=[\frac{3}{2}\ln (2x-5){]}_0^1+[-3{]}_0^1$
$=\frac{3}{2}[\ln (-3)-\ln (-5)]-3$
$=\frac{3}{2}\ln (\frac{3}{5})-3$

For ${\int }_0^{3}q(x)\cdot dx=5$, find the value of ${\int }_0^6[q(\frac{x}{2})-2x]\cdot dx$.

${\int }_0^6[q(\frac{x}{2})-2x]\cdot dx$

$={\int }_0^{6}q(\frac{x}{2})\cdot dx+{\int }_0^6-2x\cdot dx$
$=[2\cdot Q\left(\frac{x}{2}\right){]}_0^6+[-x^2{]}_0^6$
$=2\cdot [Q(3)-Q(0)]-36$
$=2\cdot 5-36$
$=-26$

For $h(x)=2x^3-x^2+2x+1$, find average value of h over the interval $[-2,1]$.

Average value is ${\int }_{-2}^{1}h(x)\cdot dx\cdot \frac{1}{3}$
$=[\frac{x^4}{2}-\frac{x^3}{3}+x^2+x{]}_{-2}^1\cdot \frac{1}{3}$
$=[\frac{1}{2}-\frac{1}{3}+1+1-(8+\frac{8}{3}+4-2)]\cdot \frac{1}{3}$
$=\frac{-7}{2}$

Find the average value of $f(x)=\cos (x)$ over the interval $[0,\frac{\pi }{6}]$.

Average value is ${\int }_0^{\frac{\pi }{6}}f(x)\cdot dx\cdot \frac{6}{\pi }$
$=[\sin (x){]}_0^{\frac{\pi }{6}}\cdot \frac{6}{\pi }$
$=\left[\sin \left(\frac{\pi }{6}\right)-\sin (0)\right]\cdot \frac{6}{\pi }$
$=[\frac{1}{2}-0]\cdot \frac{6}{\pi }$
$=\frac{3}{\pi }$

Find the average value of $f(x)=\frac{1}{3x+1}$ over the interval $[0,2]$.

Average value is ${\int }_0^{2}f(x)\cdot dx\cdot \frac{1}{2}$
$=[\frac{1}{3}\cdot \ln (3x+1){]}_0^2\cdot \frac{1}{2}$
$=[\frac{\ln (7)}{3}-\frac{\ln (1)}{3}]\cdot \frac{1}{2}$
$=\frac{\ln (7)}{3}\cdot \frac{1}{2}$
$=\frac{\ln (7)}{6}$

A quantity of gas expands according to the law $p{v}^{0.9}=75$, where $v{\text{m}}^3$ is the volume of the gas and $p{\text{Nm}}^{-2}$ is the pressure. What is the average pressure as the volume changes from $\frac{1}{2}{\text{m}}^3$ to $1{\text{m}}^3$?

We can find the function $p(v)=\frac{75}{v^{0.9}}$ so $p(v)=75\cdot v^{-0.9}$

Average pressure over the interval $v\in [\frac{1}{2},1]$ is ${\int }_{\frac{1}{2}}^{1}p(v)\cdot dv\cdot 2$
$=[75\cdot [\frac{v^{0.1}}{0.1}]{]}_{\frac{1}{2}}^1\cdot 2$
$=[750\cdot v^{0.1}{]}_{\frac{1}{2}}^1\cdot 2$
$=[750-750({\frac{1}{2}}^{0.1})]\cdot 2$
$=[1-{\frac{1}{2}}^{0.1}]\cdot 2\cdot 750$
$=1500-{\frac{1500}{2}}^{0.1}$

Therefore we found the average p over the interval.

For ${\int }_0^{5}g(x)\cdot dx=20$, find ${\int }_0^5[2g(x)+3]\cdot dx$.

${\int }_0^5[2g(x)+3]\cdot dx$
$={\int }_0^5[2g(x)]\cdot dx+{\int }_0^5[3]\cdot dx$
$=[2\cdot G(x){]}_0^5+[3x{]}_0^5$
$=2\cdot 20+15$
$=55$

Find the average value of $f(x)=\cos (x)$ over the interval $[0,\pi ]$.

Average value is ${\int }_0^{\pi }\cos (x)\cdot dx\cdot \frac{1}{\pi }$
$=[\sin (x){]}_0^{\pi }\cdot \frac{1}{\pi }$
$=0$

An object is cooling and its temperature, T degrees, after t minutes is given by $T=45e^{-2t}$. What is its average temperature over the first 15 minutes of cooling?

Answer is ${\int }_0^{15}45e^{-2t}\cdot dt\cdot \frac{1}{15}$
$=45\cdot [\frac{e^{-2t}}{-2}{]}_0^{15}\cdot \frac{1}{15}$
$=\frac{45}{-2}\cdot (e^{-30}-1)\cdot \frac{1}{15}$
$=\frac{3}{-2}\cdot (e^{-30}-1)$
$=\frac{3}{2}-\frac{3}{2}{e}^{-30}$

Find the average value of the function $f(x)=\frac{1}{3}{x}^2$ over the interval $[6,9]$.

Average value is ${\int }_6^9\frac{1}{3}{x}^2\cdot dx\cdot \frac{1}{3}$
$=\frac{1}{9}\cdot [x^3{]}_6^9\cdot \frac{1}{3}$
$=\frac{9^3-6^3}{27}$
$=19$

Find the average value of the function $f(x)=e^x+e^{-x}$ over the interval $[-2,2]$.

Average value is ${\int }_{-2}^2{e}^x+e^{-x}\cdot dx\cdot \frac{1}{4}$
$=[e^x-e^{-x}{]}_{-2}^2\cdot \frac{1}{4}$
$=[e^2-\frac{1}{e^2}-(\frac{1}{e^2}-e^2)]\cdot \frac{1}{4}$
$=[2e^2-\frac{2}{e^2}]\cdot \frac{1}{4}$
$=\frac{e^2-e^{-2}}{2}$

For $f^{\prime }(x)=g^{\prime }(x)+3$, $f(0)=3$ and $g(0)=1$. Find $f(x)$.

Find the area enclosed by the curve $y=-e^x$, both axes, and the line $x=-1$.

So the interval we want is probably $[-1.0]$

${\int }_{-1}^0-e^x\cdot dx$
$=-1\cdot [e^x{]}_{-1}^0$
$=-1(1-\frac{1}{e})$
$=\frac{1}{e}-1$

For $f(x)=x^2-3x+2$, find average value in the interval $[-2,3]$.

${\int }_{-2}^3[x^2-3x+2]\cdot dx\cdot \frac{1}{5}$
$=[\frac{x^3}{3}-\frac{3x^2}{2}+2x{]}_{-2}^3\cdot \frac{1}{5}$
$=[9-\frac{27}{2}+6-(\frac{-8}{3}-6-4)]\cdot \frac{1}{5}$
$=[25-\frac{27}{2}+\frac{8}{3}]\cdot \frac{1}{5}$
$=\frac{17}{6}$

For ${\int }_0^{a}f(x)\cdot dx=k$, find ${\int }_0^{a}3f(x)+2\cdot dx$.

${\int }_0^{a}3f(x)+2\cdot dx$
$=3k+2a$

The average value of $f(x)=x^2-2x$ over the interval $[1,a]$ is $\frac{13}{3}$. Find $a$.

${\int }_1^a[x^2-2x]\cdot dx\cdot \frac{1}{a-1}=\frac{13}{3}$
$[\frac{x^3}{3}-x^2{]}_1^a\cdot \frac{1}{a-1}=\frac{13}{3}$
$[\frac{a^3}{3}-a^2-\frac{1}{3}+1\cdot ]\frac{1}{a-1}=\frac{13}{3}$
$a=-3,5$

Given that $\cos (2x)=2{\cos }^2(x)-1$, find the value of ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\sin }^2(x)\cdot dx$.

${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\sin }^2(x)\cdot dx$.
$=$