To evaluate the given integral, we first simplify the integrand. The expression inside the integral is:
4y^2 – 5y + 12y – y^2 + y^3
Combining like terms yields:
(4y^2 – y^2) + (-5y + 12y) + y^3 = 3y^2 + 7y + y^3
Now our integral is:
∫ (y^3 + 3y^2 + 7y) dy from 1 to 2.
Next, we compute the antiderivative:
- For y^3, the antiderivative is (1/4)y^4.
- For 3y^2, the antiderivative is (1)y^3.
- For 7y, the antiderivative is (7/2)y^2.
Therefore, the antiderivative of the entire expression is:
(1/4)y^4 + y^3 + (7/2)y^2
Now we need to evaluate this from 1 to 2:
Calculating at the upper bound (y = 2):
(1/4)(2^4) + (2^3) + (7/2)(2^2) = (1/4)(16) + 8 + (7/2)(4) = 4 + 8 + 14 = 26
Now at the lower bound (y = 1):
(1/4)(1^4) + (1^3) + (7/2)(1^2) = (1/4)(1) + 1 + (7/2)(1) = 1/4 + 1 + 3.5 = 4.75
Finally, to find the value of the integral, we subtract the evaluation at the lower bound from the upper bound:
26 – 4.75 = 21.25
Thus, the value of the integral from 1 to 2 of the given expression is 21.25.