What is area y=x^2+9,y=10+2x ?

The answer to area y=x^2+9,y=10+2x is -(10sqrt(2))/3+6sqrt(2)

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area y=x^2+9,y=10+2x

area

y = x^{2} + 9, y = 10 + 2 x

- \frac{10 2}{3} + 62

Decimal Notation

3.77123 \dots

Solution steps

Apply the area formula:

\int_{1 - 2}^{1 + 2} x^{2} + 9 - (10 + 2 x) d x

Solve

\int_{1 - 2}^{1 + 2} x^{2} + 9 - (10 + 2 x) d x : - \frac{10 2}{3} + 62

The area is:

= - \frac{10 2}{3} + 62

x \to 3 lim (x^{2} - 9)

\int 3 cos (t) - 2 sin (t) d t

e^{4 t} (t^{2} + 3 t + 5)

\int (- 9 x^{2} sec (3 x^{3} + 1)) d x

\frac{d}{d x} ((4 x)^{5 x})

What is area y=x^2+9,y=10+2x ?
The answer to area y=x^2+9,y=10+2x is -(10sqrt(2))/3+6sqrt(2)