What is the integral from 3 to 4 of (7/(x^2)-1) ?

The integral from 3 to 4 of (7/(x^2)-1) is -5/12

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integral from 3 to 4 of (7/(x^2)-1)

\int_{3}^{4} (\frac{7}{x ^{2}} - 1) d x

- \frac{5}{12}

Decimal

- 0.41666 \dots

Solution steps

\int_{3}^{4} \frac{7}{x ^{2}} - 1 d x

Apply the Sum Rule:

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

= \int_{3}^{4} \frac{7}{x ^{2}} d x - \int_{3}^{4} 1 d x

\int_{3}^{4} \frac{7}{x ^{2}} d x = \frac{7}{12}

\int_{3}^{4} 1 d x = 1

= \frac{7}{12} - 1

Simplify

\frac{7}{12} - 1 : - \frac{5}{12}

= - \frac{5}{12}

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What is the integral from 3 to 4 of (7/(x^2)-1) ?
The integral from 3 to 4 of (7/(x^2)-1) is -5/12