What is area 1/4 (x-3)^2+1,x^2+1,0,3 ?

The answer to area 1/4 (x-3)^2+1,x^2+1,0,3 is 37/4

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area 1/4 (x-3)^2+1,x^2+1,0,3

area

\frac{1}{4} (x - 3)^{2} + 1, x^{2} + 1, 0, 3

\frac{37}{4}

Decimal Notation

9.25

Solution steps

Apply the area formula:

\int_{0}^{3} \frac{1}{4} (x - 3)^{2} + 1 - (x^{2} + 1) d x

Solve

\int_{0}^{3} \frac{1}{4} (x - 3)^{2} + 1 - (x^{2} + 1) d x : \frac{37}{4}

The area is:

= \frac{37}{4}

f (x) = sin^{2} (5 x)

\frac{\partial}{\partial x} (x^{3} + 2 x y)

x^{2} - 4, - 4, 3

\frac{x}{( x - 1 ) ^{2}}

n = 0 \sum \infty p^{k}

What is area 1/4 (x-3)^2+1,x^2+1,0,3 ?
The answer to area 1/4 (x-3)^2+1,x^2+1,0,3 is 37/4