What is the integral from 0 to 1 of sqrt(2-x^2) ?

The integral from 0 to 1 of sqrt(2-x^2) is pi/4+1/2

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integral from 0 to 1 of sqrt(2-x^2)

\int_{0}^{1} 2 - x^{2} d x

\frac{π}{4} + \frac{1}{2}

Decimal Notation

1.28539 \dots

Solution steps

\int_{0}^{1} 2 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 2 cos^{2} (u) d u

= \int_{0}^{\frac{π}{4}} 2 cos^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{\frac{π}{4}} cos^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + cos (2 u) d u

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d u + \int_{0}^{\frac{π}{4}} cos (2 u) d u)

\int_{0}^{\frac{π}{4}} 1 d u = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 u) d u = \frac{1}{2}

= 2 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

2 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{π}{4} + \frac{1}{2}

= \frac{π}{4} + \frac{1}{2}

\int 7 x - 1 d x

x \to 0 lim (\frac{1 + cos ( x )}{x})

\int \frac{x}{( x ^{2} + 1 ) ^{(\frac{3}{2})}} d x

n = 0 \sum \infty \frac{( 3 x ) ^{n}}{n}

x \to 3 lim (9)

What is the integral from 0 to 1 of sqrt(2-x^2) ?
The integral from 0 to 1 of sqrt(2-x^2) is pi/4+1/2