What is area x^2+4,-x^2-1,[-2,3] ?

The answer to area x^2+4,-x^2-1,[-2,3] is 145/3

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area x^2+4,-x^2-1,[-2,3]

area

x^{2} + 4, - x^{2} - 1, [- 2, 3]

\frac{145}{3}

Decimal Notation

48.33333 \dots

Solution steps

Apply the area formula:

\int_{- 2}^{3} x^{2} + 4 - (- x^{2} - 1) d x

Solve

\int_{- 2}^{3} x^{2} + 4 - (- x^{2} - 1) d x : \frac{145}{3}

The area is:

= \frac{145}{3}

\frac{d}{d x} (arcsec (x^{10}))

\frac{d}{d x} ((cos (π x - 16))^{4})

\int \frac{1}{- x ^{2} - 6 x} d x

x = - y, x = 2 y + 3, y = - 2, y = 1

\frac{d}{d x} (x ln (3))

What is area x^2+4,-x^2-1,[-2,3] ?
The answer to area x^2+4,-x^2-1,[-2,3] is 145/3