What is the integral from x^5 to x^4 of (2t-1)^3 ?

The integral from x^5 to x^4 of (2t-1)^3 is x^4(-2x^{16}+2x^{12}+4x^{11}-4x^8-3x^6+3x^4+x-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from x^5 to x^4 of (2t-1)^3

\int_{x^{5}}^{x^{4}} (2 t - 1)^{3} d t

x^{4} (- 2 x^{16} + 2 x^{12} + 4 x^{11} - 4 x^{8} - 3 x^{6} + 3 x^{4} + x - 1)

Solution steps

\int_{x^{5}}^{x^{4}} (2 t - 1)^{3} d t

Apply u-substitution

: \int_{2 x^{5} - 1}^{2 x^{4} - 1} \frac{u ^{3}}{2} d u

= \int_{2 x^{5} - 1}^{2 x^{4} - 1} \frac{u ^{3}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2 x^{5} - 1}^{2 x^{4} - 1} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{2 x^{5} - 1}^{2 x^{4} - 1}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{2 x^{5} - 1}^{2 x^{4} - 1}

Compute the boundaries:

\frac{( 2 x ^{4} - 1 ) ^{4}}{4} - \frac{( 2 x ^{5} - 1 ) ^{4}}{4}

= \frac{1}{2} (\frac{( 2 x ^{4} - 1 ) ^{4}}{4} - \frac{( 2 x ^{5} - 1 ) ^{4}}{4})

Simplify

= x^{4} (- 2 x^{16} + 2 x^{12} + 4 x^{11} - 4 x^{8} - 3 x^{6} + 3 x^{4} + x - 1)

\int_{0}^{\frac{π}{2}} - sin (x) d x

\int_{1}^{5} \frac{2}{( x - 3 ) ^{4}} d x

\int_{0}^{\frac{π}{2}} 3 sin (\frac{x}{2}) d x

\int_{1}^{2} \frac{x}{x ^{2} - 1} d x

\int_{0}^{1} 2 + x^{2} d x

What is the integral from x^5 to x^4 of (2t-1)^3 ?
The integral from x^5 to x^4 of (2t-1)^3 is x^4(-2x^{16}+2x^{12}+4x^{11}-4x^8-3x^6+3x^4+x-1)