First, let's show that f(x)=−(x−1)22−x=−(x−1)21+x−11. We can rewrite the right-hand side as a single fraction:
−(x−1)21+x−11=−(x−1)21+(x−1)2x−1=(x−1)2−1+x−1=(x−1)2x−2=−(x−1)22−x. This is equal to the given f(x). Next, we want to find K=∫−10f(x)dx=∫−10(−(x−1)21+x−11)dx. We can split this into two integrals:
K=∫−10−(x−1)21dx+∫−10x−11dx. For the first integral, let u=x−1. Then du=dx. When x=−1, u=−2. When x=0, u=−1. ∫−10−(x−1)21dx=∫−2−1−u21du=∫−2−1−u−2du=[u−1]−2−1=[u1]−2−1=−11−−21=−1+21=−21. For the second integral, let u=x−1. Then du=dx. When x=−1, u=−2. When x=0, u=−1. ∫−10x−11dx=∫−2−1u1du=[ln∣u∣]−2−1=ln∣−1∣−ln∣−2∣=ln(1)−ln(2)=0−ln(2)=−ln(2). Therefore, K=−21−ln(2).