Let's compute the integral: ∫(x·cos(x))dx For x·cos(x), we use integration by parts: ∫(u1·v2')dx = u1·v2 + ∫(u1'·v2)dx, with u1 = x and v2' = cos(x) Thus, u1' is 1 Let's compute the integral v2': ∫(cos(x))dx sin(x) Let's compute the integral u1·v2 + ∫(u1'·v2)dx: x·sin(x) -∫(sin(x))dx x·sin(x) + cos(x) Let's evaluate over the interval [0, 1]: (1·sin(1) + cos(1)) -(0·sin(0) + cos(0)) sin(1) + cos(1) + 0·sin(0) -cos(0) sin(1) + cos(1) + 0 -1 sin(1) + cos(1) -1 The approximate value of: cos(1) + sin(1) -1 is: ≈ 0.5403 + ≈ 0.8415 -1 ≈ 1.3818 -1 ≈ 0.3818 ▷Result: cos(1) + sin(1) -1 ▷▷Numeric value: ≈ 0.3818