Question

calcule a área sob a curva dada, no intervalo indicado:

A) y= x^3 de 0 a 2

b) y= sen x de 0 a π

c) y= cos^2x de 0 a π/2

Answer 1

$\mathrm{\mathbf{a)}\ y=x^3\ \ \|\ \ {[0,2]}}\\\\ \mathrm{\int\limits_0^2x^3\ dx=\dfrac{x^4}{4}\bigg|_0^2=\dfrac{2^4}{4}-\dfrac{0^4}{4}=\dfrac{16}{4}=\boxed{\mathbf{4\ u.a.}}}$


$\mathrm{\mathbf{b)}\ y=\sin{x}\ \ \|\ \ {[0,\pi]}}\\\\ \mathrm{\int\limits_0^{\pi}\sin{x}\ dx=-\cos{x}\bigg|_0^{\pi}=-\cos{\pi}-(-\cos{0})=}\\\\ \mathrm{=-(-1)+1=1+1=\boxed{\mathbf{2\ u.a.}}}$


$\mathrm{\mathbf{c)}\ y=\cos^2{x}\ \ \| \ \ {[0,\frac{\pi}{2}]}}\\\\ \mathrm{\int\limits_0^{\frac{\pi}{2}}\cos^2{x}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{1+\cos{2x}}{2}\ dx=}\\\\ \mathrm{=\dfrac{1}{2}\bigg(\int1\ dx+\int\cos{2x}\ dx\bigg)\bigg|_0^{\frac{\pi}{2}}=}\\\\ \mathrm{=\dfrac{1}{2}\bigg(x+\int\cos{2x}\ dx\bigg)\bigg|_0^{\frac{\pi}{2}}\dots}$

$\mathrm{*\ Resolvendo\ \int\cos{2x}\ dx:}\\\\ \mathrm{u=2x\ \ \|\ \ \dfrac{du}{2}=dx\ \to\ \int\cos{u}\ \dfrac{du}{2}=}\\ \mathrm{\dfrac{1}{2}\int\cos{u}\ du=\dfrac{1}{2}\sin{u}=\dfrac{\sin{2x}}{2}}$

$\mathrm{*\ Voltando\ ao\ problema:}\\\\ \mathrm{\dots\ \dfrac{1}{2}\bigg(x+\dfrac{\sin{2x}}{2}\bigg)\bigg|_0^{\frac{\pi}{2}}=}\\\\ \mathrm{=\dfrac{1}{2}\bigg(\dfrac{\pi}{2}+\dfrac{\sin{\pi}}{2}\bigg)-\dfrac{1}{2}\bigg(0+\dfrac{\sin{0}}{2}\bigg)=}\\\\ \mathrm{=\dfrac{1}{2}\bigg(\dfrac{\pi}{2}+0\bigg)=\boxed{\mathbf{\dfrac{\pi}{4}\ u.a.}}}$

Answer 2

Para a primeira temos:

$\mathtt{A = \int\limits^2_0 {x^3} \, dx \to A = \left (\dfrac{x^4}{4} \right )\limits^2_0 \to A = \dfrac{2^4}{4} - 0 \to A = 4}$

Para a segunda temos:

$\mathtt{A = \int\limits^ \pi _0 {sin(x)} \, dx \to A = \left (-cos(x) \right )\limits^{ \pi }_0 \to A =-cos( \pi ) - (-cos(0))} \\ \\ \mathtt{A = 1 + 1 \to A = 2}$

Para a terceira temos:

$\mathtt{A = \int\limits^{ \frac{ \pi }{2} }_0 {cos^2(x)} \, dx} \\ \\ \mathtt{cos^2(x) = \dfrac{1 + cos(2x)}{2} } \\ \\ \mathtt{ \int{\dfrac{1 + cos(2x)}{2} } \, dx = \dfrac{1}{2}* \int {1 + cos(2x)} \, dx = \dfrac{1}{2} (x + \dfrac{1}{2}*sin(2x)) + C} \\ \\ \mathtt{A = \int\limits^{ \frac{ \pi }{2} }_0 { \dfrac{1}{2} (x + \dfrac{1}{2}*sin(2x)) \, dx} \to A = \dfrac{1}{2} \left ( x + \dfrac{1}{2}*sin(2x)\right)\limits^{ \frac{ \pi }{2} }_0}$

$\mathtt{A = \dfrac{1}{2} \left ( \frac{ \pi }{2} + \dfrac{1}{2}*sin(2* \frac{ \pi }{2} ) \right ) - 0 \to A = \dfrac{1}{2} \left ( \frac{ \pi }{2} + \dfrac{1}{2}*sin( \pi ) \right )} \\ \\ \mathtt{A = \dfrac{ \pi }{4} - 0 \to A = \dfrac{ \pi }{4} }$