What is the integral from 1/5 to 7 of 16xln(5x) ?

The integral from 1/5 to 7 of 16xln(5x) is 16(-306/25+49/2 ln(35))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1/5 to 7 of 16xln(5x)

\int_{\frac{1}{5}}^{7} 16 x ln (5 x) d x

16 (- \frac{306}{25} + \frac{49}{2} ln (35))

Decimal Notation

1197.85644 \dots

Solution steps

\int_{\frac{1}{5}}^{7} 16 x ln (5 x) d x

Take the constant out:

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

= 16 \cdot \int_{\frac{1}{5}}^{7} x ln (5 x) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} ln (5 x) - \int \frac{x}{2} d x]_{\frac{1}{5}}^{7}

= 16 [\frac{1}{2} x^{2} ln (5 x) - \int \frac{x}{2} d x]_{\frac{1}{5}}^{7}

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= 16 [\frac{1}{2} x^{2} ln (5 x) - \frac{x ^{2}}{4}]_{\frac{1}{5}}^{7}

Compute the boundaries:

- \frac{306}{25} + \frac{49}{2} ln (35)

= 16 (- \frac{306}{25} + \frac{49}{2} ln (35))

\int_{2}^{4} (x^{2} + 4)^{3} x d x

\frac{d}{d x} (\frac{ln ( 3 x + 1 )}{6 x + 2})

\frac{9}{sin ( x ) + cos ( x )}

x, 3 x, 4 - x

\int 4 - y^{2} d y

What is the integral from 1/5 to 7 of 16xln(5x) ?
The integral from 1/5 to 7 of 16xln(5x) is 16(-306/25+49/2 ln(35))