Math  /  Calculus

QuestionFind the most general antiderivative of the function. (Check your answer by differentiation. Use CC for the constant of the antiderivative.) f(t)=2t4+7ttF(t)=\begin{array}{l} f(t)=\frac{2 t-4+7 \sqrt{t}}{\sqrt{t}} \\ F(t)=\square \end{array}

Studdy Solution

STEP 1

1. We are given the function f(t)=2t4+7tt f(t) = \frac{2t - 4 + 7\sqrt{t}}{\sqrt{t}} .
2. We need to find the most general antiderivative F(t) F(t) of f(t) f(t) .
3. The constant of integration will be denoted by C C .

STEP 2

1. Simplify the given function f(t) f(t) .
2. Integrate the simplified function to find the antiderivative.
3. Check the antiderivative by differentiating it.
4. Add the constant of integration C C .

STEP 3

Simplify the given function by dividing each term by t \sqrt{t} :
f(t)=2tt4t+7tt f(t) = \frac{2t}{\sqrt{t}} - \frac{4}{\sqrt{t}} + \frac{7\sqrt{t}}{\sqrt{t}}
f(t)=2t1/24t1/2+7 f(t) = 2t^{1/2} - 4t^{-1/2} + 7

STEP 4

Integrate each term separately:
F(t)=(2t1/24t1/2+7)dt F(t) = \int (2t^{1/2} - 4t^{-1/2} + 7) \, dt
F(t)=2t1/2dt4t1/2dt+7dt F(t) = \int 2t^{1/2} \, dt - \int 4t^{-1/2} \, dt + \int 7 \, dt

STEP 5

Calculate each integral:
1. 2t1/2dt=2t3/23/2=43t3/2\int 2t^{1/2} \, dt = 2 \cdot \frac{t^{3/2}}{3/2} = \frac{4}{3}t^{3/2}
2. 4t1/2dt=4t1/21/2=8t1/2\int 4t^{-1/2} \, dt = 4 \cdot \frac{t^{1/2}}{1/2} = 8t^{1/2}
3. 7dt=7t\int 7 \, dt = 7t

Combine the results:
F(t)=43t3/28t1/2+7t+C F(t) = \frac{4}{3}t^{3/2} - 8t^{1/2} + 7t + C

STEP 6

Differentiate F(t) F(t) to check:
ddt(43t3/28t1/2+7t+C) \frac{d}{dt}\left(\frac{4}{3}t^{3/2} - 8t^{1/2} + 7t + C\right)
=4332t1/2812t1/2+7 = \frac{4}{3} \cdot \frac{3}{2}t^{1/2} - 8 \cdot \frac{1}{2}t^{-1/2} + 7
=2t1/24t1/2+7 = 2t^{1/2} - 4t^{-1/2} + 7
This matches the original function f(t) f(t) .

STEP 7

Add the constant of integration C C :
The most general antiderivative is:
F(t)=43t3/28t1/2+7t+C F(t) = \frac{4}{3}t^{3/2} - 8t^{1/2} + 7t + C
The most general antiderivative is:
43t3/28t1/2+7t+C \boxed{\frac{4}{3}t^{3/2} - 8t^{1/2} + 7t + C}

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