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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Checkpoint 3 Worked-out solution available at LarsonAppliedCalculus.com

Evaluate 0 1 ( 2 t + 3 ) 3 d t .

To determine

To calculate: The value of definite integral 01(2t+3)3dt.

Explanation

Given Information:

The integral is 01(2t+3)3dt.

Formula used:

The fundamental theorem of calculus states that,

If f is integrable on interval [a,b] then abf(x)dx=F(b)F(a).

The integration formula is xndx=xn+1n+1+C.

Calculation:

Consider the integral,

01(2t+3)3dt

Multiply and divide by 2,

01(2t+3)3(2)dt2

First find antiderivative by applying the integration formula,

01(2t+3)3(2)dt<

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