Integrate using the simplest method possible:

∫(t – 1)5(2t – 7)dt

For me, the easiest method is to multiply everything together which will give you a sum of terms. From there you can integrate.

I think that the 5 is an exponent

∫(t – 1)^5(2t – 7)dt

expanding is still an option, but if we let u = t-1 then we have

2t-7 = 2u-5 and we have

∫u^5(2u-5)du

Now expansion is a sight less work

To integrate the given expression ∫(t – 1)^5(2t – 7)dt, we can use the method of distribution. This involves expanding the expression and then integrating each term separately.

Step 1: Expand the expression (t – 1)^5(2t – 7):

(t – 1)^5(2t – 7) = (t^5 – 5t^4 + 10t^3 – 10t^2 + 5t – 1)(2t – 7)

Step 2: Distribute the terms:

= 2t(t^5 – 5t^4 + 10t^3 – 10t^2 + 5t – 1) – 7(t^5 – 5t^4 + 10t^3 – 10t^2 + 5t – 1)

Step 3: Simplify the expression:

= 2t^6 – 10t^5 + 20t^4 – 20t^3 + 10t^2 – 2t – 7t^5 + 35t^4 – 70t^3 + 70t^2 – 35t + 7

= 2t^6 – 17t^5 + 55t^4 – 90t^3 + 80t^2 – 37t + 7

Step 4: Integrate each term separately:

∫2t^6 dt – ∫17t^5 dt + ∫55t^4 dt – ∫90t^3 dt + ∫80t^2 dt – ∫37t dt + ∫7 dt

Step 5: Use the power rule to integrate each term:

= (2/7)t^7 – (17/6)t^6 + (55/5)t^5 – (90/4)t^4 + (80/3)t^3 – (37/2)t^2 + 7t + C

Therefore, the integral of (t – 1)^5(2t – 7)dt is equal to (2/7)t^7 – (17/6)t^6 + (11/2)t^5 – (45/2)t^4 + (40/3)t^3 – (37/2)t^2 + 7t + C, where C is the constant of integration.