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

The integral from 1 to 4 of 3/(t^4) is 63/64

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integral from 1 to 4 of 3/(t^4)

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

\frac{63}{64}

Decimal Notation

0.984375

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{21}{64}

= 3 \cdot \frac{21}{64}

Simplify

= \frac{63}{64}

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

\frac{d}{d x} (8 x^{5} 2 π)

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

\frac{d}{d x} (\frac{x ^{2} + 6}{( 2 x - 7 ) ^{7}})

What is the integral from 1 to 4 of 3/(t^4) ?
The integral from 1 to 4 of 3/(t^4) is 63/64