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人気のある微分積分 >

integral from 0 to 2 of sqrt(1+(2y+1)^2)

解

\int_{0}^{2} 1 + (2 y + 1)^{2} d y

解

\frac{2 ( - 1 + 5 13 ) - ln ( 1 + 2 ) + ln ( 5 + 26 )}{4}

+1

十進法表記

6.37798 \dots

解答ステップ

\int_{0}^{2} 1 + (2 y + 1)^{2} d y

u置換積分法を適用する

= \int_{1}^{5} \frac{1 + u ^{2}}{2} d u

定数を除く:

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

= \frac{1}{2} \cdot \int_{1}^{5} 1 + u^{2} d u

三角関数による置換を適用する

= \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (5)} sec^{3} (v) d v

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (5)} sec (v) d v)

\int_{\frac{π}{4}}^{a r c t a n (5)} sec (v) d v = ln (5 + 26) - ln (1 + 2)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2))) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

限度を計算する:

\frac{5 13 - 1}{2}

= \frac{1}{2} (\frac{5 13 - 1}{2} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

簡素化

= \frac{2 ( - 1 + 5 13 ) - ln ( 1 + 2 ) + ln ( 5 + 26 )}{4}

グラフ

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