integral from-2 to 3 of 1/(1+9t^2)

解答

\int_{- 2}^{3} \frac{1}{1 + 9 t ^{2}} d t

\frac{1}{3} (arctan (9) + arctan (6))

十进制

0.95526 \dots

求解步骤

\int_{- 2}^{3} \frac{1}{1 + 9 t ^{2}} d t

使用换元积分法

= \int_{- 6}^{9} \frac{1}{3 ( 1 + u ^{2} )} d u

提出常数:

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

= \frac{1}{3} \cdot \int_{- 6}^{9} \frac{1}{1 + u ^{2}} d u

使用积分常见定则:

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

= \frac{1}{3} [arctan (u)]_{- 6}^{9}

带入上下限计算后得:

arctan (9) + arctan (6)

= \frac{1}{3} (arctan (9) + arctan (6))

a re a x = 4 y - y^{2}, x = y^{2} - 7 y

\frac{\partial}{\partial y} (6 x^{2 y})

(x^{2} - 1) y^{^{'}} + 2 x y^{2} = 0, y (0) = 1

\frac{d}{d x} (sin (x^{2} + e^{x}))

\int_{0}^{π} 3 sin (x) d x