Definite integrals
Problem 4.11 · easy
Evaluate \( \displaystyle \int_{0}^{\frac{\pi}{2}} 2 \sin{\left(x \right)} + 2 \cos{\left(x \right)} \, dx \).
- By the Fundamental Theorem of Calculus, Part 2, the integral is F(b) − F(a) for any antiderivative F.
- \[ \frac{d}{d x} \left(2 \sin{\left(x \right)} - 2 \cos{\left(x \right)}\right) = 2 \sin{\left(x \right)} + 2 \cos{\left(x \right)} \]F(x) = 2*sin(x) - 2*cos(x) is an antiderivative: its derivative is the integrand.✓ Proved
- \[ - \left. 2 \sin{\left(x \right)} - 2 \cos{\left(x \right)} \right|_{\substack{ x=0 }} + \left. 2 \sin{\left(x \right)} - 2 \cos{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{2} }} = 4 \]F(b) − F(a).✓ Proved
Answer \( 4 \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | numerical quadrature (mpmath) gives the same value |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/definite_integral, checked 2026-09-26 with SymPy 1.14.0.