integral from 0 to x of sqrt(x^2+y^2)

解

\int_{0}^{x} x^{2} + y^{2} d y

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

解答ステップ

\int_{0}^{x} x^{2} + y^{2} d y

三角関数による置換を適用する

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

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

= x^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

簡素化

x^{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : x^{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= x^{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

限度を計算する:

\frac{1}{2}

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

簡素化

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