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

解

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

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

解答ステップ

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

三角関数による置換を適用する

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

定数を除く:

\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

積分換算を適用する

= 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)))

簡素化

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)))

限度を計算する:

2

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