ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 5 to 10 of sqrt(x^2+5)

解

\int_{5}^{10} x^{2} + 5 d x

解

5 (105 - \frac{15}{2} + \frac{1}{2} (ln (25 + 21) - ln (5 + 6)))

+1

十進法表記

39.18869 \dots

解答ステップ

\int_{5}^{10} x^{2} + 5 d x

三角関数による置換を適用する

= \int_{a r c t a n (5)}^{a r c t a n (25)} 5 sec^{3} (u) d u

定数を除く:

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

= 5 \cdot \int_{a r c t a n (5)}^{a r c t a n (25)} sec^{3} (u) d u

積分換算を適用する

= 5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (25)} + \frac{1}{2} \cdot \int_{a r c t a n (5)}^{a r c t a n (25)} sec (u) d u

\int_{a r c t a n (5)}^{a r c t a n (25)} sec (u) d u = ln (25 + 21) - ln (5 + 6)

= 5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (25)} + \frac{1}{2} (ln (25 + 21) - ln (5 + 6))

簡素化

5 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (5)}^{a r c t a n (25)} + \frac{1}{2} (ln (25 + 21) - ln (5 + 6)) : 5 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (5)}^{a r c t a n (25)} + \frac{1}{2} (ln (25 + 21) - ln (5 + 6)))

= 5 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (5)}^{a r c t a n (25)} + \frac{1}{2} (ln (25 + 21) - ln (5 + 6)))

限度を計算する:

105 - \frac{15}{2}

= 5 (105 - \frac{15}{2} + \frac{1}{2} (ln (25 + 21) - ln (5 + 6)))

グラフ

人気の例

integral from 0 to 8 of 9x^2+3

\int_{0}^{8} 9 x^{2} + 3 d x

integral from 0 to 1 of x(1-x^2)^{1/2}

\int_{0}^{1} x (1 - x^{2})^{\frac{1}{2}} d x

integral from 0 to 100 of 9x

\int_{0}^{100} 9 x d x

integral from-1 to 1 of xln(x+2)

\int_{- 1}^{1} x ln (x + 2) d x

integral from 1.7 to 4.7 of 2t^{1/2}

\int_{1.7}^{4.7} 2 t^{\frac{1}{2}} d t