What is the integral from 1 to 2 of 3/t ?

The integral from 1 to 2 of 3/t is 3ln(2)

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integral from 1 to 2 of 3/t

\int_{1}^{2} \frac{3}{t} d t

3 ln (2)

Decimal Notation

2.07944 \dots

Solution steps

\int_{1}^{2} \frac{3}{t} d t

Take the constant out:

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

= 3 \cdot \int_{1}^{2} \frac{1}{t} d t

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= 3 [ln ∣ t ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= 3 ln (2)

y = x^{2} - 3 x, (1, - 2)

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 cos (x)

t y^{^{'}} + (t + 1) y = t, y (ln (2)) = 1

\int (6 sin (2 x) + sec^{2} (x)) d x

What is the integral from 1 to 2 of 3/t ?
The integral from 1 to 2 of 3/t is 3ln(2)