What is area 3x^2+12,4x+4,[-3,3] ?

The answer to area 3x^2+12,4x+4,[-3,3] is 102

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area 3x^2+12,4x+4,[-3,3]

area

3 x^{2} + 12, 4 x + 4, [- 3, 3]

102

Solution steps

Apply the area formula:

\int_{- 3}^{3} 3 x^{2} + 12 - (4 x + 4) d x

Solve

\int_{- 3}^{3} 3 x^{2} + 12 - (4 x + 4) d x : 102

The area is:

= 102

\frac{1}{1 + x + x ^{2} + x ^{3}}

\frac{d}{d x} (\frac{18}{( 1 + 3 x ) ^{3}})

\frac{25 s}{( s ^{2} + 9 ) ( s + 4 )}

1 + sin^{2} (x)

y^{^{'}} - \frac{2}{t} y = t^{2}

What is area 3x^2+12,4x+4,[-3,3] ?
The answer to area 3x^2+12,4x+4,[-3,3] is 102