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人気のある微分積分 >

integral from-2 to 2 of sqrt(x^2+4)

解

\int_{- 2}^{2} x^{2} + 4 d x

解

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

+1

十進法表記

9.18234 \dots

解答ステップ

\int_{- 2}^{2} x^{2} + 4 d x

三角関数による置換を適用する

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

定数を除く:

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

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

積分換算を適用する

= 4 ([\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)

= 4 ([\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

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

= 4 ([\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

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

グラフ

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