FIND THE AREA UNDER THE CURVE
-3X2SQUARE+X+14
2007-02-22 04:05:43 · 4 answers · asked by Djkid 1 in Science & Mathematics ➔ Mathematics
CAN SOME ONE SOLVE THIS FOR ME ? THNX
2007-02-22 04:13:47 · update #1
The corect answer is that you integrate it and put the upper and lower limits. But if u need some help with integration heere it is
-3X^3(cube)/3 +x^2(square)/2 +14 *x
so -x^3+x^2+14x and then just sub in the value of x and u are good to go.
2007-02-22 04:14:55 · answer #1 · answered by UnENG 3 · 0⤊ 0⤋
INTEGRATION
-3x^2 + x + 14 = 0
Find roots where f(x)=0
(-3x + 7) (x + 2) = 0
x=-2 and x = 7/3
integrate
-3/3x^3 +1/2 x^2 + 14X + C = 0
Area = S f(7/3) - Sf(-2)
Sf(7/3) = 22.685
Sf(-2) = -18
Area = 40.685
-(17/3)^3 + 1/2(7/3)^2 + 14(7/3)
-(-2)^3 + 1/2(-2)^2 + 14(-2)
2007-02-22 12:20:10 · answer #2 · answered by Grant d 4 · 0⤊ 0⤋
-3x^2 +x+14 has roots of 7/3 and -2.
A = integ -2 to 7/3 [(-3x^2 +x +14)dx]
A = -x^3 +x^2/2 +14x evaluated from -2 to 7/3
A =(-(7/3)^3 + (7/3)^2/2 +(7/3)(14) -(-2)^3+ (-2)^2/2 +14(-2))
A = (-12.70 +2.72 +32.66) -(-8+2 -28)
A= 22.66 + 34
A= 56.66
2007-02-22 12:22:46 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
just use integration, and substitute the upper and lower limits into x
2007-02-22 12:08:46 · answer #4 · answered by Ramani 2 · 0⤊ 0⤋