Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2 (e) Determine, if it exists, a real number a < 2 such that fe² f(x) dx = 2 e² f(x) dx. We have to solve the equation in the unknown a, given by (E) L(x²-6 6x+8)e dx == Integrating by parts, we get the following primitives Therefore (x²-6x (x2 - 6x+8)e* dx. 6r+8)e dr=8-6 re-edr+ze-2fre de dx = 8e-6xe + 6e + x²e - 2xe + 2e+c e [x28x+16]+c=e(x-4)²+c, CER. why not e-z & why a negative sign I understand this Integration by parts twice f(x) dr=4² - (0-4)² I do not understand how the limits [f(x) dr = 4². → were substituted Therefore equation (E) is equivalent to 4e2 - ea (a-4)² = 4e2, or -ea (a - 4)² = 0, whose only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2. Therefore the equation has no solutions. Exercise 1. Consider the function f: dom(f) CRR defined as f(x) = x²-6x+8|ex-2. (e) Determine, if it exists, a real number a < 2 such that fe² f(x) da = 2 e² f(x) dx. We have to solve the equation in the unknown a, given by (E) L(x²-6 6r+8) dr == (r²-6x+8)e* dr. 12 why not e-z & why a negative sign Integrating by parts, we get the following primitives Therefore (x²-6x 6x+8)e dx = 8e6 -6-d]+-2/dr e dx + x2e-2 xe dr 8e6xe+ 6e + x²e - 2xe + 2e+c e [x2-8x+16]+c=e(x-4)²+c, CER. = I understand this Integration by parts twice f(r) dr-42-(0-4)² I do not understand how the limits f(x) dr=42 → were substituted Therefore equation (E) is equivalent to 4e2 - ea (a - 4)² = 4e², or -ea (a - 4)² = 0, whose only solution is a = 4. We cannot accept this solution, since we gave the condition a < 2. Therefore the equation has no solutions.