10. $\int_{1}^{4} t^{-3 / 2} d t=$
(A) -1
(B) $-\frac{7}{8}$
(C) $-\frac{1}{2}$
(D) $\frac{1}{2}$
(E) 1

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We need to evaluate the definite integral $ \int_{1}^{4} t^{-3/2} \, dt $. ‖ ‖ step 2 ⋮ The antiderivative of $ t^{-3/2} $ is $ \int t^{-3/2} \, dt = -2 t^{-1/2} + C $. ‖ ‖ step 3 ⋮ We now evaluate this antiderivative from 1 to 4: $ \left[ -2 t^{-1/2} \right]_{1}^{4} $. ‖ ‖ step 4 ⋮ Substituting the limits, we get $ -2 (4^{-1/2}) - (-2 (1^{-1/2})) = -2 \left( \frac{1}{2} \right) - (-2 \cdot 1) $. ‖ ‖ step 5 ⋮ Simplifying, we get $ -1 + 2 = 1 $. ‖ ***E*** ∻Key Concept∻ ⚹Definite Integral⚹ ∻Explanation∻ ⚹The definite integral of a function over an interval can be found by evaluating the antiderivative at the endpoints of the interval and subtracting.⚹◊What is the value of the definite integral $\int_{1}^{4} t^{-3/2} dt$⍭ Generate me a similar question◊

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