What is area 4x=y^2+12,3<= x<= 11 ?

The answer to area 4x=y^2+12,3<= x<= 11 is (128sqrt(2))/3

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area 4x=y^2+12,3<= x<= 11

area

4 x = y^{2} + 12, 3 \leq x \leq 11

\frac{128 2}{3}

Decimal Notation

60.33977 \dots

Solution steps

Isolate

y

for

4 x = y^{2} + 12 : y = 2 x - 3, y = - 2 x - 3

Apply the area formula:

\int_{3}^{11} 4 x - 3 d x

Solve

\int_{3}^{11} 4 x - 3 d x : \frac{128 2}{3}

The area is:

= \frac{128 2}{3}

\int_{0}^{\frac{π}{2}} ∣ 7 sin (x) - 7 cos (2 x) ∣ d x

y = \frac{2}{x ^{2}}

\int_{6}^{\infty} \frac{1}{x ^{2} + x} d x

\frac{d y}{d t} = 2 - y

\frac{d y}{d x} = x^{4} (32 - x^{5})^{5}

What is area 4x=y^2+12,3<= x<= 11 ?
The answer to area 4x=y^2+12,3<= x<= 11 is (128sqrt(2))/3