1. Julieta vasquez escuela 72 ci:22202274Hallar el rea:x-4 = x-4x-x = 4-4x-x = 0x (x-1) = 0x=0 o x-1 =0x=0 o x=1tablas de valoresF(x)= x-4x 0 0,5 1F(x) -4 -3,75 -3G(x) x-4x 0 0,5 1G(x) -4 -3,5 -31( x4( x-4) )dxArea = 01( x4x2+4 ) dx= 01( xx2 )dx= 0 2. = (x22 x33 )10x22 x33 -0Area= 16u B) y= xy= 4xx=4xx-4x=0x=(x-2) (x+2) = 0x=0 o x=2 o x=-2Construymos la tabla de valores :Y=xx -2 -1 0 1 2y -8 -1 0 1 8Y=4xx -2 0 2y -8 0 8Ahora la grafica: 3. Deacuerdo a la grafica:0 ( x34 x) dx + 0Area: 22 (4 xx3) dxArea = ( x44 2 x )+ (2 x x44 )=04(2 ) 44 - (2(0)-2(-2))+(2(2) - 2(0)) 2 4044= (-4+8) + (8-4)Area= 8uC) x=12y x=0 y=1 y=eConstruyamos la tablax 12 6 3 1,6y 1 2 4 e 4. Asie ( 12y ) d yArea= 1= (12 L N / y/ ) e 1= 12 LN (e) 12 LN /1/= 12(2) 12(0)Area= 24 uD) F(x) = tan ( x2 ) ; eje X , x=0 , x=2Tabla de valores:x 0 42F(x) 0 0,41 1 5. Asi2tan( xArea: 02 ) dxTomando: u= x2 du=d x2 2du=dxX= 0 => U=0X=2=> U= 4Resulta :4tan(u) d uArea 02 6. =124tan(u)d u0= c o s( x )/12(l n )= -12ln(c o s( 4 )lnco s(0 )/= -12ln /12 / - ln 1Area=12ln( 2 ) u