Question

1- Calcule:

2- Encontre a área limitada pelo eixo x e o gráfico de

=

2 − , com 0 ≤ ≤ 2.

3- Calcule a área do conjunto de todos os pontos (, )

tais que

2 ≤ ≤ √.

4- Responda:​

Answer 1

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1)

$\tt a)~\displaystyle\sf\int_0^1\sqrt{1+x^2}dx\\\sf x=1tg~\theta\implies dx=sec^2~\theta\\\sf se~x=0\implies \theta=0\\\sf se~x=1\implies \theta=\frac{\pi}{4}\\\sf \sqrt{1+x^2}=\sqrt{1+tg^2\theta}=\sqrt{sec^2\theta}=sec~\theta$

$\displaystyle\sf\int_0^1\sqrt{1+x^2}dx=\int_0^{\frac{\pi}{4}}sec~\theta\cdot sec^2~\theta~d\theta=\int_0^{\frac{\pi}{4}}sec^3~\theta~d\theta\\\boxed{\displaystyle\sf\int sec^3~\theta~d\theta=\dfrac{1}{2}\left[sec~\theta\cdot tg~\theta+\ell n|sec~\theta+tg~\theta|\right]+k}\\\displaystyle\sf \int_0^{\frac{\pi}{4}}sec^3~\theta~d\theta=\dfrac{1}{2}\left[sec~\theta\cdot tg~\theta+\ell n|sec~\theta+tg~\theta|\right]\Bigg|_{0}^{\frac{\pi}{4}}$

$\sf sec\left(\dfrac{\pi}{4}\right)=\sqrt{2}~~tg\left(\dfrac{\pi}{4}\right)=1\\\sf sec(0)=1~tg(0)=0\\\displaystyle\sf\int_{0}^{\frac{\pi}{4}}sec^3~\theta~d\theta\\\sf=\dfrac{1}{2}\left[sec\left(\dfrac{\pi}{4}\right)\cdot tg\left(\dfrac{\pi}{4}\right)+\ell n|sec\left(\dfrac{\pi}{4}\right)+tg\left(\dfrac{\pi}{4}\right)|\right-\left\{sec(0)\cdot tg(0)+\ell n|sec(0)+tg(0)|\right\}]\\\boxed{\displaystyle\sf\int_{0}^{\frac{\pi}{4}} sec^3~\theta~d\theta=\dfrac{1}{2}\left[\sqrt{2}\cdot1+\ell n|\sqrt{2}+1|]}$$\tt b)\\\displaystyle\sf\int x^3\sqrt{1+x^2}~dx\\\sf u=1+x^2\implies x^2=u-1\\\sf du=2x~dx\implies x~dx=\dfrac{1}{2}du\\\sf x^3dx=x^2\cdot x~dx=(u-1)\dfrac{1}{2}du\\\displaystyle\sf \int x^3\sqrt{1+x^2}dx=\int\sqrt{u}(u-1)\dfrac{1}{2}~du=\dfrac{1}{2}\int(u^{\frac{3}{2}}-u^{\frac{1}{2}})du\\\sf =\dfrac{1}{\diagup\!\!\!2}\cdot\dfrac{\diagup\!\!\!2}{5}u^{\frac{5}{\diagup\!\!\!2}}-\dfrac{1}{\diagup\!\!\!2}\cdot\dfrac{\diagup\!\!\!2}{3}u^{\frac{3}{2}}+k$

$\boxed{\displaystyle\sf\int x^3\sqrt{1+x^2}dx=\dfrac{1}{5}(1+x^2)^{\frac{5}{2}}-\dfrac{1}{3}(1+x^2)^{\frac{3}{2}}+k}$

integral que produz função trigonométrica inversa

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int\dfrac{du}{a^2+u^2}=\dfrac{1}{a}~arctg\left(\dfrac{u}{a}\right)+k}}}}$

$\tt c)\\\displaystyle\sf\int\dfrac{5}{4+x^2}dx=5\int\dfrac{dx}{2^2+x^2}=5\cdot\dfrac{1}{2}arctg\left(\dfrac{x}{2}\right)+k\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int\dfrac{5}{4+x^2}dx=\dfrac{5}{2}~arctg\left(\dfrac{x}{2}\right)+k}}}}$

Relação fundamental da trigonometria

$\huge\boxed{\boxed{\boxed{\boxed{\sf sen^2x+cos^2x=1}}}}$

$\tt d)~\displaystyle\sf\int sen^3(3x)\cdot cos^3(3x)~dx=\int sen^3(3x)\cdot cos^2(3x)\cdot cos(3x)~dx\\\sf sen^2(3x)+cos^2(3x)=1\implies cos^2(3x)=1-sen^2(3x)\\\displaystyle\sf\int sen^3(3x)\cdot(1-sen^2(3x))\cdot cos(3x)~dx \\\sf fac_{\!\!,}a\\\sf u=sen(3x)\implies\dfrac{1}{3}du=cos(3x)~dx\\\displaystyle\sf\int sen^3(3x)\cdot(1-sen^2(3x)\cdot cos(3x)~dx=\dfrac{1}{3}\int u^3(1-u^2)du$

$\displaystyle\sf\int u^3(1-u^2)~du=\int(u^3-u^5)du=\dfrac{1}{4}u^4-\dfrac{1}{6}u^6+k\\\displaystyle\sf\dfrac{1}{3}(u^3-u^5)du=\dfrac{1}{3}\cdot\dfrac{1}{4}u^4-\dfrac{1}{3}\cdot\dfrac{1}{6}u^6+k=\dfrac{1}{12}u^4-\dfrac{1}{18}u^6+k\\\boxed{\displaystyle\sf\int sen^3(3x)\cdot cos^3(3x)~dx=-\dfrac{1}{12}sen^4(3x)-\dfrac{1}{18}sen^6(3x)+k}$

$\tt e)\\\displaystyle\sf\int\dfrac{x+3}{x^2-3x+2}dx\\\sf x^2-3x+2=(x-1)(x-2)\\\sf\dfrac{x+3}{x^2-3x+2}=\dfrac{x+3}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}\\\sf A=\dfrac{x+3}{x-2}\Bigg|_{x=1}=\dfrac{1+3}{1-2}=-4\\\sf B=\dfrac{x+3}{x-1}\Bigg|_{x=2}=\dfrac{2+3}{2-1}=5$

$\displaystyle\sf\int\dfrac{x+3}{x^2-3x+2}dx=\int\dfrac{x+3}{(x-1)(x-2)}dx=-4\int\dfrac{dx}{x-1}}+5\int\dfrac{dx}{x-2}\\\displaystyle\sf\int\dfrac{x+3}{(x-1)(x-2)}dx=-4\ell n|x-1|+5\ell n|x-2|+k\\\displaystyle\sf\int\dfrac{dx}{(x-1)(x-2)}=\ell n\left|\dfrac{(x-2)^5}{(x-1)^4}\right|+k\\\boxed{\displaystyle\sf\int\dfrac{x+3}{x^2-3x+2}dx=\ell n\left|\dfrac{(x-2)^5}{(x-1)^4}\right|+k}\checkmark$

2)

$\displaystyle\sf\int_0^2( x^2-x)~dx=\int_0^1(0-[x^2-x])dx+\int_1^2(x^2-x)dx\\\displaystyle\sf\int_0^1(x-x^2)dx=\dfrac{1}{2}x^2-\dfrac{1}{3}x^3\Bigg|_{0}^1\\\sf =\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}\\\displaystyle\sf\int_1^2(x^2-x)dx=\dfrac{1}{3}x^3-\dfrac{1}{2}x^2\Bigg|_{1}^2\\\sf =\dfrac{1}{3}2^3-\dfrac{1}{2}2^2-\left[\dfrac{1}{3}-\dfrac{1}{2}\right]=\dfrac{8}{3}-\dfrac{4}{2}-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{16-12-2+3}{6}=\dfrac{5}{6}$

$\displaystyle\sf\int_0^2(x^2-x)dx=\dfrac{1}{6}+\dfrac{5}{6}=1$

3)

$\displaystyle\sf A=\int_0^1(\sqrt{x}-x^2)dx=\dfrac{1}{2}x^{\frac{3}{2}}-\dfrac{1}{3}x^3\Bigg|_{0}^1=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}~u\cdot a$

4)

$\tt a)~\sf uma~primitiva~representa~o~contr\acute ario~da~derivada.\\\tt b)~\sf calcula~a~\acute area~sob~a~curva.$