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

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

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

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

x^{2} (2 + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2)))

Solution steps

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{4}}^{\frac{π}{4}} x^{2} sec^{3} (u) d u

= \int_{- \frac{π}{4}}^{\frac{π}{4}} x^{2} sec^{3} (u) d u

Take the constant out:

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

= x^{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= x^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sec (u) d u)

\int_{- \frac{π}{4}}^{\frac{π}{4}} sec (u) d u = ln (1 + 2) - ln (- 1 + 2)

= x^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2)))

Simplify

x^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))) : x^{2} ([\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2)))

= x^{2} ([\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2)))

Compute the boundaries:

2

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

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