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人気のある微分積分 >

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

解

\int_{1}^{3} x^{2} + 1 d x

解

\frac{2 ( - 1 + 3 5 ) - ln ( 1 + 2 ) + ln ( 3 + 10 )}{2}

+1

十進法表記

4.50484 \dots

解答ステップ

\int_{1}^{3} x^{2} + 1 d x

三角関数による置換を適用する

= \int_{\frac{π}{4}}^{a r c t a n (3)} sec^{3} (u) d u

積分換算を適用する

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (3)} sec (u) d u

\int_{\frac{π}{4}}^{a r c t a n (3)} sec (u) d u = ln (3 + 10) - ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

簡素化

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

限度を計算する:

\frac{3 5 - 1}{2}

= \frac{3 5 - 1}{2} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

簡素化

= \frac{2 ( - 1 + 3 5 ) - ln ( 1 + 2 ) + ln ( 3 + 10 )}{2}

グラフ

人気の例

integral from 1 to 2 of x^2-1/2

\int_{1}^{2} x^{2} - \frac{1}{2} d x

integral from 2 to 3 of (2x-x^2)

\int_{2}^{3} (2 x - x^{2}) d x

integral from t to 0 of e^{ax}

\int_{t}^{0} e^{a x} d x

integral from 0 to 4 of pi|4x-x^2|

\int_{0}^{4} π 4 x - x^{2} d x

integral from-pi/4 to pi/4 of x^3sin(x)

\int_{- \frac{π}{4}}^{\frac{π}{4}} x^{3} sin (x) d x