What is the integral from 1/2 to 3/4 of 1/(u^2) ?

The integral from 1/2 to 3/4 of 1/(u^2) is 2/3

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integral from 1/2 to 3/4 of 1/(u^2)

\int_{\frac{1}{2}}^{\frac{3}{4}} \frac{1}{u ^{2}} d u

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{\frac{1}{2}}^{\frac{3}{4}} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{\frac{1}{2}}^{\frac{3}{4}}

= [- \frac{1}{u}]_{\frac{1}{2}}^{\frac{3}{4}}

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

\int_{- 1}^{3} 3 x^{2} + 4 x d x

y = arcsec (x^{2} + 1)

y = ln (x^{3} - 2 x^{2} + 2)

x \to \infty lim (\frac{2}{x} + 1)

\int 10 sin (t) + 3 cos (t) d t

What is the integral from 1/2 to 3/4 of 1/(u^2) ?
The integral from 1/2 to 3/4 of 1/(u^2) is 2/3