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

The integral from 1 to 5 of sqrt(1+(3)^2) is 4sqrt(10)

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integral from 1 to 5 of sqrt(1+(3)^2)

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

410

Decimal Notation

12.64911 \dots

Solution steps

\int_{1}^{5} 1 + 3^{2} d x

Integral of a constant:

\int a d x = a x

= [1 + 3^{2} x]_{1}^{5}

Simplify

= [10 x]_{1}^{5}

Compute the boundaries:

410

= 410

\int_{0}^{t} e^{- 2 t} sin (5 t) d t

\frac{d}{d x} (9 cos^{3} (x))

\int_{0}^{\frac{1}{6}} x ln (6 x) d x

y = x + 1, (- 1, 4)

x \to \infty lim (i^{x})

What is the integral from 1 to 5 of sqrt(1+(3)^2) ?
The integral from 1 to 5 of sqrt(1+(3)^2) is 4sqrt(10)