What is the integral from 1 to x of 3t^{-4} ?

The integral from 1 to x of 3t^{-4} is -1/(x^3)+1

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integral from 1 to x of 3t^{-4}

\int_{1}^{x} 3 t^{- 4} d t

- \frac{1}{x ^{3}} + 1

Solution steps

\int_{1}^{x} 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}^{x} t^{- 4} d t

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{3 x ^{3}} + \frac{1}{3}

= 3 (- \frac{1}{3 x ^{3}} + \frac{1}{3})

Simplify

= - \frac{1}{x ^{3}} + 1

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\int_{2}^{3} 5 x^{4} - 8 x^{3} + 1 d x

\int_{0}^{1} - 2 x + 3 d x

What is the integral from 1 to x of 3t^{-4} ?
The integral from 1 to x of 3t^{-4} is -1/(x^3)+1