What is the integral from x to 1 of ysqrt(1+x^2-y^2) ?

The integral from x to 1 of ysqrt(1+x^2-y^2) is -(x^3-1)/3

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integral from x to 1 of ysqrt(1+x^2-y^2)

\int_{x}^{1} y 1 + x^{2} - y^{2} d y

- \frac{x ^{3} - 1}{3}

Solution steps

\int_{x}^{1} y 1 + x^{2} - y^{2} d y

Apply u-substitution

: \int_{1}^{x^{2}} - \frac{u}{2} d u

= \int_{1}^{x^{2}} - \frac{u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int_{1}^{x^{2}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{x^{2}}

= - \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{x^{2}}

Compute the boundaries:

\frac{2}{3} x^{3} - \frac{2}{3}

= - \frac{1}{2} (\frac{2}{3} x^{3} - \frac{2}{3})

Simplify

= - \frac{x ^{3} - 1}{3}

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