What is area y=(10)/(x^2-4x-21),x=-2,x=2 ?

The answer to area y=(10)/(x^2-4x-21),x=-2,x=2 is 2ln(3)

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area y=(10)/(x^2-4x-21),x=-2,x=2

area

y = \frac{10}{x ^{2} - 4 x - 21}, x = - 2, x = 2

2 ln (3)

Decimal

2.19722 \dots

Solution steps

Apply the area formula:

\int_{- 2}^{2} \frac{10}{∣ x ^{2} - 4 x - 21 ∣} d x

Solve

\int_{- 2}^{2} \frac{10}{∣ x ^{2} - 4 x - 21 ∣} d x : 2 ln (3)

The area is:

= 2 ln (3)

g (x) = x^{3} - 6 x^{2} + 9 x + 3

x = 4 - y^{2}, x = y^{2} - 4

\int \frac{p}{( p - 5 ) ^{2}} d p

\frac{d y}{d x} = 2 y + x^{2} + 9

\frac{d y}{d t} (t) = \frac{y}{25 t ^{2} + 1}

What is area y=(10)/(x^2-4x-21),x=-2,x=2 ?
The answer to area y=(10)/(x^2-4x-21),x=-2,x=2 is 2ln(3)