Evaluate the Integral

Problem

(∫_0^1)((2*t−1)2*d(t))

Solution

1. Expand the integrand by squaring the binomial (2*t−1)2

(2*t−1)2=4*t2−4*t+1

2. Apply the power rule for integration, (∫_^)(tn*d(t))=(t(n+1))/(n+1) to each term of the polynomial.

(∫_^)((4*t2−4*t+1)*d(t))=(4*t3)/3−(4*t2)/2+t

3. Simplify the expression before evaluating the definite integral.

(4*t3)/3−2*t2+t

4. Evaluate the definite integral by plugging in the upper limit 1 and the lower limit 0

[(4*(1)3)/3−2*(1)2+1]−[(4*(0)3)/3−2*(0)2+0]

5. Calculate the final numerical value.

4/3−2+1=4/3−1=1/3

Final Answer

(∫_0^1)((2*t−1)2*d(t))=1/3

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