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

The integral from-4 to-1 of-4/(x^3) is 15/8

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

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

\frac{15}{8}

Decimal

1.875

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{- 4}^{- 1}

= - 4 [- \frac{1}{2 x ^{2}}]_{- 4}^{- 1}

Compute the boundaries:

- \frac{15}{32}

= - 4 (- \frac{15}{32})

Simplify

= \frac{15}{8}

y^{^{'}} + 3 y = e^{t}

f (x) = 110 x - 16 x^{2}, at

y = ln e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x)

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

(7, - 3), (11, - 4)

What is the integral from-4 to-1 of-4/(x^3) ?
The integral from-4 to-1 of-4/(x^3) is 15/8