What is the integral from 2 to 4 of (1-(x-3)^2) ?

The integral from 2 to 4 of (1-(x-3)^2) is 4/3

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integral from 2 to 4 of (1-(x-3)^2)

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Expand

1 - (x - 3)^{2} : - x^{2} + 6 x - 8

= \int_{2}^{4} - x^{2} + 6 x - 8 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{2}^{4} x^{2} d x + \int_{2}^{4} 6 x d x - \int_{2}^{4} 8 d x

\int_{2}^{4} x^{2} d x = \frac{56}{3}

\int_{2}^{4} 6 x d x = 36

\int_{2}^{4} 8 d x = 16

= - \frac{56}{3} + 36 - 16

Simplify

- \frac{56}{3} + 36 - 16 : \frac{4}{3}

= \frac{4}{3}

\frac{\partial}{\partial x} (x + 1)

\int_{1}^{4} y ln (y) d y

n = 0 \sum \infty (\frac{e}{2})^{n}

\int x ln (x + 3) d x

\int_{1}^{e} 8 t^{2} ln (t) d t

What is the integral from 2 to 4 of (1-(x-3)^2) ?
The integral from 2 to 4 of (1-(x-3)^2) is 4/3