Question

Nos problemas seguintes, calcule a integral definida dada usando o teorema
fundamental do cálculo.

Answer 1

Respostas de cada item:

• a) 3π
• b) 0
• c) 28/3
• d) 1/3
• e) 7/2

⠀

Pelo Teorema Fundamental do Cálculo, tem-se que:

$\displaystyle\int^b_af(x)~dx=F(x)\Big|^b_a=F(b)-F(a)$

Onde F(x) é a primitiva de f(x), dentro do intervalo variando de a até b.

Em integrais definidas, geralmente calculamos a integral indefinida para depois por em prática esse teorema.

⠀

Item a)

$\displaystyle\int\limits^1_{-\,2}\pi~dx$

Não há muito o que fazer aqui; por ser a integral de uma constante temos:

$\displaystyle\int\pi~dx=\pi x+c$

Sendo πx a primitiva de π e ''c'' a constante, pensando na família de derivadas (lembrando que desconsideramos a constante ao calcular a integral definida). Daí, pelo teorema:

$\displaystyle\int\limits^1_{-\,2}\pi~dx=\pi x\Big|^1_{-\,2}=\pi\cdot1-\pi\cdot(-\,2)$

$\displaystyle\int\limits^1_{-\,2}\pi~dx=\pi+2\pi$

$\boxed{\displaystyle\int\limits^1_{-\,2}\pi~dx=3\pi}$

⠀

Item b)

$\displaystyle\int\limits^4_15-2t~dt$

Pelas propriedades:

$\displaystyle\int5-2t~dt$

$\displaystyle\int5~dt-\displaystyle\int2t~dt$

$\displaystyle\int5~dt-2\displaystyle\int t~dt$

$5t+c_1-\dfrac{2t^2}{2}+c_2$

$5t-t^2+C$

Daí, pelo teorema:

$\displaystyle\int\limits^4_15-2t~dt=5t-t^2\Big|^4_1=5\cdot4-4^2-(5\cdot1-1^2)$

$\displaystyle\int\limits^4_15-2t~dt=20-16-(5-1)$

$\displaystyle\int\limits^4_15-2t~dt=4-4$

$\boxed{\displaystyle\int\limits^4_15-2t~dt=0}$

⠀

Item c)

$\displaystyle\int\limits^4_12\sqrt{u}~du$

Aplicando a propriedade:

$\displaystyle\int2\sqrt{u}~du$

$2\displaystyle\int\sqrt{u}~du$

$2\displaystyle\int u^{\frac{1}{2}}~du$

Pela regra da potência, tem-se:

$2\cdot\dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+2c$

$2\cdot\dfrac{u^{\frac{1+2}{2}}}{\frac{1+2}{2}}+2c$

$2\cdot\dfrac{u^{\frac{3}{2}}}{\frac{3}{2}}+2c$

$\dfrac{4u^{\frac{3}{2}}}{3}+2c$

Aplicando o teorema:

$\text{ \displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{4u^{\frac{3}{2}}}{3}\:\Bigg|^4_1=\dfrac{4\cdot4^{\frac{3}{2}}}{3}-\dfrac{4\cdot1^{\frac{3}{2}}}{3} }$

$\displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{4^{\frac{3}{2}+1}}{3}-\dfrac{4\cdot1}{3}$

$\displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{4^{\frac{3+2}{2}}}{3}-\dfrac{4}{3}$

$\displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{(2^2)^\frac{5}{2}-4}{3}$

$\displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{2^{\frac{2\cdot5}{2}}-4}{3}$

$\displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{2^5-4}{3}$

$\displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{32-4}{3}$

$\boxed{\displaystyle\int\limits^4_12\sqrt{u}~du=\dfrac{28}{3}}$

⠀

Item d)

$\displaystyle\int\limits^9_4x^{-\frac{3}{2}}~dx$

Aqui não há o que se fazer se não aplicar a regra da potência:

$\dfrac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+c$

$\dfrac{x^{-\frac{3+2}{2}}}{-\frac{3+2}{2}}+c$

$\dfrac{x^{-\frac{1}{2}}}{-\frac{1}{2}}+c$

$-\dfrac{2}{x^\frac{1}{2}}+c$

Daí, pelo teorema:

$\text{ \displaystyle\int\limits^9_4x^{-\frac{3}{2}}~dx=-\dfrac{2}{x^\frac{1}{2}}\Bigg|^9_4=-\dfrac{2}{9^\frac{1}{2}}-\bigg(\!\!-\dfrac{2}{4^\frac{1}{2}}\bigg) }$

$\displaystyle\int\limits^9_4x^{-\frac{3}{2}}~dx=-\dfrac{2}{(3^2)^\frac{1}{2}}+\dfrac{2}{(2^2)^\frac{1}{2}}$

$\displaystyle\int\limits^9_4x^{-\frac{3}{2}}~dx=-\dfrac{2}{3}+\dfrac{2}{2}$

$\displaystyle\int\limits^9_4x^{-\frac{3}{2}}~dx=-\dfrac{2}{3}+\dfrac{3}{3}$

$\displaystyle\int\limits^9_4x^{-\frac{3}{2}}~dx=-\dfrac{2+3}{3}$

$\boxed{\displaystyle\int\limits^9_4x^{-\frac{3}{2}}~dx=\dfrac{1}{3}}$

⠀

Item e)

$\displaystyle\int\limits^0_{-\,1}-\,3x^5-3x^2+2x+5~dx$

Pela propriedade:

$\displaystyle\int-\,3x^5-3x^2+2x+5~dx$

$\displaystyle\int-\,3x^5~dx-\displaystyle\int3x^2~dx+\displaystyle\int2x~dx+\displaystyle\int5~dx$

$-\,3\displaystyle\int x^5~dx-3\displaystyle\int x^2~dx+2\displaystyle\int x~dx+\displaystyle\int5~dx$

$-\dfrac{3x^6}{6}+c_1-\dfrac{3x^3}{3}+c_2+\dfrac{2x^2}{2}+c_3+5x+c_4$

$-\dfrac{x^6}{2}-x^3+x^2+5x+C$

E por fim, aplicando a teorema:

$\displaystyle\int\limits^0_{-\,1}-\,3x^5-3x^2+2x+5~dx=-\dfrac{x^6}{2}-x^3+x^2+5x\Bigg|^0_{-\,1}$$=-\dfrac{0^6}{2}-0^3+0^2+5\cdot0-\bigg(\!\!-\dfrac{(-\,1)^6}{2}-(-\,1)^3+(-\,1)^2+5\cdot(-\,1)\bigg)$

$\displaystyle\int\limits^0_{-\,1}-\,3x^5-3x^2+2x+5~dx=-\dfrac{0}{2}-0+0+0+\dfrac{(1)}{2}+(-\,1)-(1)-(-\,5)$

$\displaystyle\int\limits^0_{-\,1}-\,3x^5-3x^2+2x+5~dx=0+\dfrac{1}{2}-1-1+5$

$\displaystyle\int\limits^0_{-\,1}-\,3x^5-3x^2+2x+5~dx=\dfrac{1}{2}+3$

$\displaystyle\int\limits^0_{-\,1}-\,3x^5-3x^2+2x+5~dx=\dfrac{1+6}{2}$

$\boxed{\displaystyle\int\limits^0_{-\,1}-\,3x^5-3x^2+2x+5~dx=\dfrac{7}{2}}$

Bons estudos e um forte abraço. — lordCzarnian9635.