What is the integral from 0 to 1 of x^2(1+4x^3)^5 ?

The integral from 0 to 1 of x^2(1+4x^3)^5 is 217

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of x^2(1+4x^3)^5

\int_{0}^{1} x^{2} (1 + 4 x^{3})^{5} d x

217

Solution steps

\int_{0}^{1} x^{2} (1 + 4 x^{3})^{5} d x

Apply u-substitution

: \int_{1}^{5} \frac{u ^{5}}{12} d u

= \int_{1}^{5} \frac{u ^{5}}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{1}^{5} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{1}^{5}

= \frac{1}{12} [\frac{u ^{6}}{6}]_{1}^{5}

Compute the boundaries:

2604

= \frac{1}{12} \cdot 2604

Simplify

= 217

\int_{1}^{3} x^{- 1} + x^{2} d x

\int_{0}^{1} \frac{1}{4} x^{2} d x

\int_{0}^{x} 6 + 7 t^{2} d t

\int_{0}^{x} ∣ x ∣ d x

\int_{0}^{\infty} y x e^{- x y} d y

What is the integral from 0 to 1 of x^2(1+4x^3)^5 ?
The integral from 0 to 1 of x^2(1+4x^3)^5 is 217