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人気のある微分積分 >

integral from-1 to 3 of sqrt(1+(2y+1)^2)

解

\int_{- 1}^{3} 1 + (2 y + 1)^{2} d y

解

\frac{1}{2} (182 + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

+1

十進法表記

13.60929 \dots

解答ステップ

\int_{- 1}^{3} 1 + (2 y + 1)^{2} d y

u置換積分法を適用する

= \int_{- 1}^{7} \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}^{7} 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 (7)} 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 (7)} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{a r c t a n (7)} sec (v) d v)

\int_{- \frac{π}{4}}^{a r c t a n (7)} sec (v) d v = ln (7 + 52) - ln (- 1 + 2)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} (ln (7 + 52) - 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 (7)} + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2))) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

限度を計算する:

182

= \frac{1}{2} (182 + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

グラフ

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