What is area y=(x^2)/2+2,[-3,0] ?

The answer to area y=(x^2)/2+2,[-3,0] is 21/2

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area y=(x^2)/2+2,[-3,0]

area

y = \frac{x ^{2}}{2} + 2, [- 3, 0]

\frac{21}{2}

Decimal

10.5

Solution steps

Apply the area formula:

\int_{- 3}^{0} \frac{x ^{2}}{2} + 2 d x

Solve

\int_{- 3}^{0} \frac{x ^{2}}{2} + 2 d x : \frac{21}{2}

The area is:

= \frac{21}{2}

\frac{d}{d x} ((3 x + 6)^{2})

\frac{d}{d x} (- \frac{11}{x ^{12}})

y = - x^{2} - 2 x + 3, y = - x + 1

\int \frac{( x - 1 )}{( x ^{2} - 2 x - 35 ) ^{7}} d x

\frac{9}{t}