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人気のある微分積分 >

integral from-2 to 2 of sqrt(1+100x^2)

解

\int_{- 2}^{2} 1 + 100 x^{2} d x

解

\frac{1}{10} (20401 + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

+1

十進法表記

40.41891 \dots

解答ステップ

\int_{- 2}^{2} 1 + 100 x^{2} d x

三角関数による置換を適用する

= \int_{- a r c t a n (20)}^{a r c t a n (20)} \frac{1}{10} sec^{3} (u) d u

定数を除く:

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

= \frac{1}{10} \cdot \int_{- a r c t a n (20)}^{a r c t a n (20)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{10} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} \cdot \int_{- a r c t a n (20)}^{a r c t a n (20)} sec (u) d u)

\int_{- a r c t a n (20)}^{a r c t a n (20)} sec (u) d u = ln (20 + 401) - ln (- 20 + 401)

= \frac{1}{10} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

簡素化

\frac{1}{10} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401))) : \frac{1}{10} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

= \frac{1}{10} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (20)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

限度を計算する:

20401

= \frac{1}{10} (20401 + \frac{1}{2} (ln (20 + 401) - ln (- 20 + 401)))

グラフ

人気の例

f (x) = - cot (x)

integral of (3x^3+6x^2-12x+5)

\int (3 x^{3} + 6 x^{2} - 12 x + 5) d x

derivative f(x)=t^5sin(t)

d er i v a t i v e f (x) = t^{5} sin (t)

integral from 0 to 1 of sqrt(1+144x)

\int_{0}^{1} 1 + 144 x d x

derivative t^2sqrt(t)

d er i v a t i v e t^{2} t