Question

INTEGRAL Calcular a área:

1) Y = 1 + 2X intervalo x=3 e x=1

2) Y = X2 + 3X + 2 intervalo x=2 e x=0


3) Y = 5 – X intervalo x= 2 e x=0

Answer 1

De forma geral temos que:

$\int_a^{b} f(x)\; dx = F(x)]_a^{b}=F(b)-F(a)$

Onde F(x) é a primitiva de f(x)

1

$\int_1^{3} 2x+1\; dx = 2\int_1^{3} x\; dx+\int_1^{3} 1\; dx$

Resolvemos cada integral separadamente

$2\int_1^{3} x\; dx=2\left ( \dfrac{x^2}{2} \right ) \Bigg]_1^{3} =2\left ( \dfrac{3^2}{2}-\dfrac{1^2}{2} \right ) =8$

$\int_1^{3} 1\; dx=1x]_1^{3}=(1\cdot3)-(1\cdot1)=2$

Por fim temos:

$\int_1^{3} 2x+1\; dx = 2\int_1^{3} x\; dx+\int_1^{3} 1\; dx=8+2=10\; u.a.$

2

$\int_0^{2}x^2+ 3x+2\; dx = \int_0^{2}x^2\; dx+3\int_0^{2} x\; dx+\int_0^{2} 2\; dx$

Resolvemos cada integral separadamente

$\int_0^{2} x^2\; dx= \dfrac{x^3}{3} \Bigg]_0^{2}=\dfrac{2^3}{3}-\dfrac{0^3}{3}=\frac{8}{3}$

$3\int_0^{2} x\; dx=3\left ( \dfrac{x^2}{2} \right ) \Bigg]_0^{2} =3\left ( \dfrac{2^2}{2}-\dfrac{0^2}{2} \right ) =6$

$\int_0^{2} 2\; dx=1x]_0^{2} =(2\cdot2)-(2\cdot0)=4$

Por fim temos:

$\int_0^{2}x^2+ 3x+2\; dx = \int_0^{2}x^2\; dx+3\int_0^{2} x\; dx+\int_0^{2} 2\; dx=\dfrac{8}{3}+6+4=\dfrac{38}{3}\; u.a.$

3

$\int_0^{2}5-x\; dx = \int_0^{2}5\; dx-\int_0^{2} x\; dx$

Resolvemos cada integral separadamente

$\int_0^{2} 5\; dx=5x]_0^{2} =(5\cdot2)-(5\cdot0)=10$

$\int_0^{2} x\; dx=\left ( \dfrac{x^2}{2} \right ) \Bigg]_0^{2} =\left ( \dfrac{2^2}{2}-\dfrac{0^2}{2} \right ) =2$

Por fim temos:

$\int_0^{2}5-x\; dx = \int_0^{2}5\; dx-\int_0^{2} x\; dx=10-2=8\; u.a.$