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人気のある微分積分 >

integral from 0 to 1/3 of 3/(1+9x^2)

解

\int_{0}^{\frac{1}{3}} \frac{3}{1 + 9 x ^{2}} d x

解

\frac{π}{4}

+1

十進法表記

0.78539 \dots

解答ステップ

\int_{0}^{\frac{1}{3}} \frac{3}{1 + 9 x ^{2}} d x

定数を除く:

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

= 3 \cdot \int_{0}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

置換積分法を適用する

= 3 \cdot \int_{0}^{1} \frac{1}{3 ( 1 + u ^{2} )} d u

定数を除く:

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

= 3 \cdot \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

共通積分を使用する:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 3 \cdot \frac{1}{3} [arctan (u)]_{0}^{1}

簡素化

3 \cdot \frac{1}{3} [arctan (u)]_{0}^{1} : [arctan (u)]_{0}^{1}

= [arctan (u)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= \frac{π}{4}

グラフ

人気の例

integral from-5 to 1 of |x+3|

\int_{- 5}^{1} ∣ x + 3 ∣ d x

(\partial ^2)/(\partial y^2)(x^2+y^2)

\frac{\partial ^{2}}{\partial y ^{2}} (x^{2} + y^{2})

y^'+9x^{-1}y=x^7,y(1)=8

y^{^{'}} + 9 x^{- 1} y = x^{7}, y (1) = 8

integral of (5x-3)^2(2x-1)

\int (5 x - 3)^{2} (2 x - 1) d x

limit as x approaches 0 of xln(1+1/x)

x \to 0 lim (x ln (1 + \frac{1}{x}))