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

The integral from 0 to 1 of sqrt(4+4x^2+x^4) is 7/3

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

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

\frac{7}{3}

Decimal Notation

2.33333 \dots

Solution steps

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

Factor

4 + 4 x^{2} + x^{4} : (x^{2} + 2)^{2}

= \int_{0}^{1} (x^{2} + 2)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} (x^{2} + 2)^{2} d x

Eliminate Absolutes

= \int_{0}^{1} (x^{2} + 2)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{1} (x^{2} + 2) d x

Eliminate Absolutes

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

Apply the Sum Rule:

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

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

= \frac{1}{3} + 2

Simplify

\frac{1}{3} + 2 : \frac{7}{3}

= \frac{7}{3}

\int_{e}^{1} e^{x^{2}} d x

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

\int_{0}^{2} 7 x + 1 d x

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

\int_{- 1}^{1} 2 x^{2} - x + 6 d x

What is the integral from 0 to 1 of sqrt(4+4x^2+x^4) ?
The integral from 0 to 1 of sqrt(4+4x^2+x^4) is 7/3