What is the integral from 2 to 6 of 2400xsqrt(3+2x) ?

The integral from 2 to 6 of 2400xsqrt(3+2x) is -3360sqrt(7)+36000sqrt(15)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- 33607 + 3600015

Decimal

130537.67605 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{7}^{15} \frac{u ( u - 3 )}{4} d u

= 2400 \cdot \int_{7}^{15} \frac{u ( u - 3 )}{4} d u

Take the constant out:

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

= 2400 \cdot \frac{1}{4} \cdot \int_{7}^{15} u (u - 3) d u

Expand

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

= 2400 \cdot \frac{1}{4} \cdot \int_{7}^{15} u^{\frac{3}{2}} - 3 u d u

Apply the Sum Rule:

\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_{7}^{15} u^{\frac{3}{2}} d u - \int_{7}^{15} 3 u d u)

\int_{7}^{15} u^{\frac{3}{2}} d u = 9015 - \frac{98 7}{5}

\int_{7}^{15} 3 u d u = 3015 - 147

= 2400 \cdot \frac{1}{4} (9015 - \frac{98 7}{5} - (3015 - 147))

Simplify

2400 \cdot \frac{1}{4} (9015 - \frac{98 7}{5} - (3015 - 147)) : - 33607 + 3600015

= - 33607 + 3600015

\int_{1}^{2} 4 x^{3} - 6 x d x

\int_{0}^{2} 3^{x} - 4 d x

\int_{0}^{x} x^{2} + \frac{2}{y ^{2}} d y

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

\int_{1}^{3} (\frac{15}{26} x^{2} - 2) d x

What is the integral from 2 to 6 of 2400xsqrt(3+2x) ?
The integral from 2 to 6 of 2400xsqrt(3+2x) is -3360sqrt(7)+36000sqrt(15)