What is area 4-(x-2)^2,5-(x-4)^2,2,4 ?

The answer to area 4-(x-2)^2,5-(x-4)^2,2,4 is 17/4

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area 4-(x-2)^2,5-(x-4)^2,2,4

area

4 - (x - 2)^{2}, 5 - (x - 4)^{2}, 2, 4

\frac{17}{4}

Decimal Notation

4.25

Solution steps

Apply the area formula:

\int_{2}^{4} 4 - (x - 2)^{2} - (5 - (x - 4)^{2}) d x

Solve

\int_{2}^{4} 4 - (x - 2)^{2} - (5 - (x - 4)^{2}) d x : \frac{17}{4}

The area is:

= \frac{17}{4}

\int_{0}^{2} (- x + 2)^{2} d x

\int 4 cos^{2} (2 x) + 4 sin^{2} (2 x) d x

\frac{d}{d y} (e^{x^{2} - y^{2}})

\int - 7 x^{3} (6 x^{4} + 1)^{6} d x

x \to - \infty lim (\frac{1}{4 x})

What is area 4-(x-2)^2,5-(x-4)^2,2,4 ?
The answer to area 4-(x-2)^2,5-(x-4)^2,2,4 is 17/4