What is the integral from 1 to 4 of (2-sqrt(t))/(t^2) ?

The integral from 1 to 4 of (2-sqrt(t))/(t^2) is 1/2

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integral from 1 to 4 of (2-sqrt(t))/(t^2)

\int_{1}^{4} \frac{2 - t}{t ^{2}} d t

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{1}^{4} \frac{2 - t}{t ^{2}} d t

Expand

\frac{2 - t}{t ^{2}} : \frac{2}{t ^{2}} - \frac{1}{t ^{\frac{3}{2}}}

= \int_{1}^{4} \frac{2}{t ^{2}} - \frac{1}{t ^{\frac{3}{2}}} d t

Apply the Sum Rule:

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

= \int_{1}^{4} \frac{2}{t ^{2}} d t - \int_{1}^{4} \frac{1}{t ^{\frac{3}{2}}} d t

\int_{1}^{4} \frac{2}{t ^{2}} d t = \frac{3}{2}

\int_{1}^{4} \frac{1}{t ^{\frac{3}{2}}} d t = 1

= \frac{3}{2} - 1

Simplify

\frac{3}{2} - 1 : \frac{1}{2}

= \frac{1}{2}

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What is the integral from 1 to 4 of (2-sqrt(t))/(t^2) ?
The integral from 1 to 4 of (2-sqrt(t))/(t^2) is 1/2