zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral from 6 to 9 of 2400xsqrt(3+2x)

解答

\int_{6}^{9} 2400 x 3 + 2 x d x

解答

8064021 - 3600015

+1

十进制

230111.50357 \dots

求解步骤

\int_{6}^{9} 2400 x 3 + 2 x d x

提出常数:

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

= 2400 \cdot \int_{6}^{9} x 3 + 2 x d x

使用换元积分法

= 2400 \cdot \int_{15}^{21} \frac{u ( u - 3 )}{4} d u

提出常数:

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

= 2400 \cdot \frac{1}{4} \cdot \int_{15}^{21} u (u - 3) d u

乘开

u (u - 3) : u^{\frac{3}{2}} - 3 u

= 2400 \cdot \frac{1}{4} \cdot \int_{15}^{21} u^{\frac{3}{2}} - 3 u d u

使用积分加法定则:

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

= 2400 \cdot \frac{1}{4} (\int_{15}^{21} u^{\frac{3}{2}} d u - \int_{15}^{21} 3 u d u)

\int_{15}^{21} u^{\frac{3}{2}} d u = \frac{882 21}{5} - 9015

\int_{15}^{21} 3 u d u = 3 (1421 - 1015)

= 2400 \cdot \frac{1}{4} (\frac{882 21}{5} - 9015 - 3 (1421 - 1015))

化简

2400 \cdot \frac{1}{4} (\frac{882 21}{5} - 9015 - 3 (1421 - 1015)) : 8064021 - 3600015

= 8064021 - 3600015

作图

流行的例子

integral from 0 to pi of (cos(θ))^2

\int_{0}^{π} (cos (θ))^{2} d θ

integral from 1 to 2 of 1/(xsqrt(ln(x)))

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

integral from 0 to 8 of-2t+5

\int_{0}^{8} - 2 t + 5 d t

integral from 7 to 3 of (3x(x^2-1)^3)/4

\int_{7}^{3} \frac{3 x ( x ^{2} - 1 ) ^{3}}{4} d x

integral from 0 to 50 of (10-0.2x)

\int_{0}^{50} (10 - 0.2 x) d x