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

The integral from 1 to 2 of 7/(x^2) is 7/2

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

\int_{1}^{2} \frac{7}{x ^{2}} d x

\frac{7}{2}

Decimal Notation

3.5

Solution steps

\int_{1}^{2} \frac{7}{x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int_{1}^{2} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{2}

= 7 [- \frac{1}{x}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= 7 \cdot \frac{1}{2}

Simplify

= \frac{7}{2}

\frac{d y}{d x} = 3 x^{- 2} e^{- y}

\int \frac{( x ^{2} + 9 x + 1 )}{( x ^{2} + 1 ) ^{2}} d x

x^{2} y^{^{''}} + 4 x y^{^{'}} - 4 y = 0, y (1) = 0, y^{^{'}} (1) = 3

\int \frac{225}{x ^{3} - 15 x ^{2}} d x

\int \frac{x ^{2} + 17}{x} d x

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