Consider the functionand let be a positive real number. For , let be a point on the curve . Let denote the value of such that the distance between the foot of the perpendicular from to the -axis and the point where the tangent to the curve at meets the -axis is equal to . Evaluate[4 points]\subsection*{Numerical answer}If the value is , compute .\
[THOUGHT] The distance described is the vertical distance along the y-axis between (0,f(s)) and the y-intercept of the tangent at x=s. Since f'(x)=x^2/(1+x)>0 for s>0, this distance equals s f'(s)=s^3/(1+s). Thus g(t) is the inverse of t(s)=s^3/(1+s), and \int g(t)dt from t=a to b becomes \int_{s(a)}^{s(b)} s * (dt/ds) ds. I will find s-values for t=1/2 and 27/4 (they turn out to be 1 and 3), then compute the exact integral with Sympy and extract p,q from p/q+ln2. [/THOUGHT] [PYTHON] import sympy as sp
s=sp.symbols('s', positive=True) t_expr = s**3/(1+s) dt_ds = sp.diff(t_expr, s) integrand = s*dt_ds sp.factor(integrand)
endpoints
sol1 = sp.solve(sp.Eq(t_expr, sp.Rational(1,2)), s) sol2 = sp.solve(sp.Eq(t_expr, sp.Rational(27,4)), s) print("solutions t=1/2:", sol1) print("solutions t=27/4:", sol2)
choose positive real root
s1=sp.Integer(1) s2=sp.Integer(3)
I = sp.integrate(integrand, (s, s1, s2)) I_simplified = sp.simplify(I) print("Integral:", I_simplified)
express as rational + log(2)
I_log_simplified = sp.nsimplify(I_simplified, [sp.log(2)]) print("nsimplify:", I_log_simplified)
extract p/q if form p/q + log(2)
expr = sp.simplify(I_log_simplified - sp.log(2)) rat = sp.Rational(expr) print("p/q:", rat, "p+q:", rat.p + rat.q) print("numeric check:", float(I_simplified.evalf())) [/PYTHON][THOUGHT] The distance described is the vertical distance along the y-axis between (0,f(s)) and the y-intercept of the tangent at x=s. Since f'(x)=x^2/(1+x)>0 for s>0, this distance equals s f'(s)=s^3/(1+s). Thus g(t) is the inverse of t(s)=s^3/(1+s), and \int g(t)dt from t=a to b becomes \int_{s(a)}^{s(b)} s * (dt/ds) ds. I will find s-values for t=1/2 and 27/4 (they turn out to be 1 and 3), then compute the exact integral with Sympy and extract p,q from p/q+ln2. [/THOUGHT] [PYTHON] import sympy as sp
s=sp.symbols('s', positive=True) t_expr = s**3/(1+s) dt_ds = sp.diff(t_expr, s) integrand = sp.factor(s*dt_ds)
endpoints
sol1 = sp.solve(sp.Eq(t_expr, sp.Rational(1,2)), s) sol2 = sp.solve(sp.Eq(t_expr, sp.Rational(27,4)), s) print("solutions t=1/2:", sol1) print("solutions t=27/4:", sol2)
s1=sp.Integer(1) s2=sp.Integer(3)
I = sp.integrate(integrand, (s, s1, s2)) I_simplified = sp.simplify(I) print("integrand:", integrand) print("Integral:", I_simplified)
express as rational + log(2)
I_log_simplified = sp.nsimplify(I_simplified, [sp.log(2)]) print("nsimplify:", I_log_simplified)
extract p/q if form p/q + log(2)
expr = sp.simplify(I_log_simplified - sp.log(2)) rat = sp.Rational(expr) print("p/q:", rat, "p+q:", rat.p + rat.q)
print("numeric check:", float(I_simplified.evalf())) [/PYTHON]
[THOUGHT] The integral equals (\log(2)+\frac{157}{12}), so (p+q=157+12=169). \boxed{169}