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

The integral from 0 to 2 of ()/(sqrt(4+t^2)) is (2dt)/(sqrt(t^2+4))

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integral from 0 to 2 of ()/(sqrt(4+t^2))

\int_{0}^{2} \frac{d t}{4 + t ^{2}} d x

\frac{2 d t}{t ^{2} + 4}

Solution steps

\int_{0}^{2} \frac{d t}{4 + t ^{2}} d x

Integral of a constant:

\int a d x = a x

= [\frac{d t}{4 + t ^{2}} x]_{0}^{2}

Compute the boundaries:

2 \cdot \frac{d t}{t ^{2} + 4}

= 2 \cdot \frac{d t}{t ^{2} + 4}

Simplify

= \frac{2 d t}{t ^{2} + 4}

y^{^{''}} + 49 y = 0, y (0) = 4, y^{^{'}} (0) = - 6

y^{^{'}} - y = 3 x - 3

f (x) = 1 - \frac{1}{x + 2}

- \frac{3}{2} x^{2} + 3 x + 3

\int \frac{1}{r} d r

What is the integral from 0 to 2 of ()/(sqrt(4+t^2)) ?
The integral from 0 to 2 of ()/(sqrt(4+t^2)) is (2dt)/(sqrt(t^2+4))