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人気のある微分積分 >

integral from 0 to 1 of x^2sqrt(1+4x^2)

解

\int_{0}^{1} x^{2} 1 + 4 x^{2} d x

解

\frac{- ln ( 2 + 5 ) + 18 5}{64}

+1

十進法表記

0.60633 \dots

解答ステップ

\int_{0}^{1} x^{2} 1 + 4 x^{2} d x

三角関数による置換を適用する

= \int_{0}^{a r c t a n (2)} \frac{1}{8} sec^{3} (u) tan^{2} (u) d u

定数を除く:

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

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) tan^{2} (u) d u

三角関数の公式を使用して書き換える

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) (- 1 + sec^{2} (u)) d u

拡張

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (2)} - sec^{3} (u) + sec^{5} (u) d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (- \int_{0}^{a r c t a n (2)} sec^{3} (u) d u + \int_{0}^{a r c t a n (2)} sec^{5} (u) d u)

\int_{0}^{a r c t a n (2)} sec^{3} (u) d u = 5 + \frac{1}{2} ln (2 + 5)

\int_{0}^{a r c t a n (2)} sec^{5} (u) d u = \frac{13 5 + \frac{3}{2} ln ( 2 + 5 )}{4}

= \frac{1}{8} (- (5 + \frac{1}{2} ln (2 + 5)) + \frac{13 5 + \frac{3}{2} ln ( 2 + 5 )}{4})

簡素化

\frac{1}{8} (- (5 + \frac{1}{2} ln (2 + 5)) + \frac{13 5 + \frac{3}{2} ln ( 2 + 5 )}{4}) : \frac{- ln ( 2 + 5 ) + 18 5}{64}

= \frac{- ln ( 2 + 5 ) + 18 5}{64}

グラフ

人気の例

integral from 0 to 5 of 5ln(x)

\int_{0}^{5} 5 ln (x) d x

integral from-3 to 3 of 1/(x^2)

\int_{- 3}^{3} \frac{1}{x ^{2}} d x

integral from 0 to 2pi of cos^4(t)

\int_{0}^{2 π} cos^{4} (t) d t

integral from 0 to pi/4 of cos(2t)

\int_{0}^{\frac{π}{4}} cos (2 t) d t

integral from-pi/2 to pi/2 of (cos(x)-5)

\int_{- \frac{π}{2}}^{\frac{π}{2}} (cos (x) - 5) d x