## Thomas' Calculus 13th Edition

a. The average value is defined as: $\dfrac{1}{2} \int_{-1}^1 (mx+b ) dx$ $\dfrac{1}{2} [(m)\dfrac{x^{1+1}}{1+1}+bx]_{-1}^1=\dfrac{1}{2} [m(\dfrac{1}{2} -\dfrac{1}{2} )+b(1+1)]=b$ b. The average value is defined as $\dfrac{1}{2k} \int_{-k}^k (mx+b ) dx$ Then, $\dfrac{1}{2k} [(m)\dfrac{x^{1+1}}{1+1}+bx]_{-k}^k1=\dfrac{1}{2k} [m(\dfrac{k}{2}-\dfrac{k}{2})+b(k+k)]=(\dfrac{1}{2k}) (2bk)=b$