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

The integral from 1 to 2 of 3/(x^4) is 7/8

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

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

\frac{7}{8}

Decimal Notation

0.875

Solution steps

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

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}{x ^{4}} d x

Apply the Power Rule

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

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

Compute the boundaries:

\frac{7}{24}

= 3 \cdot \frac{7}{24}

Simplify

= \frac{7}{8}

\frac{d r}{d θ} = \frac{r ^{2} + 4 r θ}{2 θ ^{2}}

f (x) = x^{4} + 3 x - 2, (1, 2)

x \to 0 - lim (2^{\frac{1}{x}})

sin (3 x^{2} + 2)

\int \frac{2 x - 1}{x ^{2} + 1} d x

What is the integral from 1 to 2 of 3/(x^4) ?
The integral from 1 to 2 of 3/(x^4) is 7/8