What is the integral from 1 to t of 35x^{-3} ?

The integral from 1 to t of 35x^{-3} is 35(-1/(2t^2)+1/2)

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integral from 1 to t of 35x^{-3}

\int_{1}^{t} 35 x^{- 3} d x

35 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

Solution steps

\int_{1}^{t} 35 x^{- 3} d x

Take the constant out:

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

= 35 \cdot \int_{1}^{t} x^{- 3} d x

Apply the Power Rule

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

= 35 [- \frac{1}{2 x ^{2}}]_{1}^{t}

Compute the boundaries:

- \frac{1}{2 t ^{2}} + \frac{1}{2}

= 35 (- \frac{1}{2 t ^{2}} + \frac{1}{2})

\int \frac{x ^{3}}{3 ( x ^{2} + 1 )} d x

e^{t a n (x)}

f (x) = \frac{x}{x ^{2} + 64}

2 (1 - e^{- t})

n = 1 \sum \infty n (\frac{3}{4})^{n}

What is the integral from 1 to t of 35x^{-3} ?
The integral from 1 to t of 35x^{-3} is 35(-1/(2t^2)+1/2)