Question

Alguém poderia me ajudar??? 55 Pontos!
a) Determine a área da região limitada pelo gráfico da função f(x) = 4x +3, pelas retas x = −1 e x = 4 e pelo eixo x. É Faça um esboço do gráfico.
3 e y = √x
b) Achar a área entre a curva x
c) Determine a área da parte positiva da área limitada pelas curvas y = x3 e y = 4x.

Answer 1

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$\tt a)\\\underline{\rm vamos~dividir~a~\acute area~em~duas~regi\tilde oes:}\\\sf regi\tilde ao~1: de~-1~a~-\dfrac{3}{4}\\\sf regi\tilde ao~2: de~-\dfrac{3}{4}~a~4.\\\displaystyle\sf A_1=\int_{-1}^{-\frac{3}{4}}(0-[4x+3])~dx\\\displaystyle\sf A_1=\int_{-1}^{-\frac{3}{4}}(-4x-3)~dx=\bigg[-2x^2-3x\bigg]_{-1}^{-\frac{3}{4}}\\\sf A_1=-2\cdot\bigg(-\dfrac{3}{4}\bigg)^2-3\cdot\bigg(-\dfrac{3}{4}\bigg)-[-2\cdot(-1)^2-3\cdot(-1)]\\\sf A_1=-\dfrac{9}{8}+\dfrac{9}{4}-[-2+3]\\\sf A_1=-\dfrac{9}{8}+\dfrac{9}{4}-1$

$\sf A_1=\dfrac{-9+18-8}{8}\\\sf A_1=\dfrac{18-17}{8}\\\sf A_1=\dfrac{1}{8}$

$\displaystyle\sf A_2=\int_{-\frac{3}{4}}^4(4x+3)~dx=\bigg[2x^2+3x\bigg]_{-\frac{3}{4}}^4\\\sf A_2=2\cdot4^2+3\cdot4-\bigg[2\cdot\bigg(-\dfrac{3}{4}\bigg)^2+3\cdot\bigg(-\dfrac{3}{4}\bigg)\bigg]\\\sf A_2=32+12-\bigg[\dfrac{9}{8}-\dfrac{9}{4}\bigg]\\\sf A_2=44-\dfrac{9}{8}+\dfrac{9}{4}\\\sf A_2=\dfrac{352-9+18}{8}\\\sf A_2=\dfrac{361}{8}$

$\sf A_{tota\ell}=A_1+A_2\\\sf A_{tota\ell}=\dfrac{1}{8}+\dfrac{361}{8}\\\sf A_{tota\ell}=\dfrac{362\div2}{8\div2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A_{tota\ell}=\dfrac{181}{4}}}}}\blue{\checkmark}$

$\tt b)\\\displaystyle\sf A=\int_0^1(\sqrt{x}-x^3)~dx=\bigg[\dfrac{2}{3}x^{\frac{3}{2}}-\dfrac{1}{4}x^4\bigg]_0^1\\\sf A=\dfrac{2}{3}\cdot1^{\frac{3}{2}}-\dfrac{1}{4}\cdot1^4\\\sf A=\dfrac{2}{3}-\dfrac{1}{4}\\\sf A=\dfrac{8-3}{12}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{5}{12}~u\cdot a}}}}\blue{\checkmark}$

$\tt c)\\\displaystyle\sf A=\int_0^2(4x-x^3)~dx=\bigg[2x^2-\dfrac{1}{4}x^4\bigg]_0^2\\\sf A=2\cdot2^2-\dfrac{1}{4}\cdot2^4\\\sf A=8-4\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=4~u\cdot a}}}}\blue{\checkmark}$