An antiderivative F(x) of f(x) satisfies: — F'(x) = f(x).An antiderivative satisfies F'(x) = f(x) — differentiating the antiderivative returns the original function.
Why is there a +C in indefinite integrals? — Because the derivative of a constant is 0, so infinitely many functions share the same derivative.The derivative of any constant is 0. So F(x) + C and F(x) + K have the same derivative f(x). The constant C represents this ambiguity.
∫x³ dx = ? — x⁴/4 + C.∫xⁿ dx = xⁿ⁺¹/(n+1) + C. ∫x³ dx = x⁴/4 + C.
∫eˣ dx = ? — eˣ + C.eˣ is its own antiderivative: ∫eˣ dx = eˣ + C.
∫sin x dx = ? — −cos x + C.∫sin x dx = −cos x + C. (Verify: d/dx[−cos x] = sin x. ✓)
∫cos x dx = ? — sin x + C.∫cos x dx = sin x + C. (Verify: d/dx[sin x] = cos x. ✓)
∫(3x² + 2x − 1) dx = ? — Both B and C.∫3x² dx = x³, ∫2x dx = x², ∫1 dx = x. Total: x³ + x² − x + C. Options B and C express the same answer.
∫5 dx = ? — 5x + C.∫k dx = kx + C. ∫5 dx = 5x + C.
Verify ∫2x dx = x² + C by: — Differentiating x² + C to get 2x.Always verify antiderivatives by differentiating: d/dx[x² + C] = 2x. ✓
What is the general formula for ∫xⁿ dx (n ≠ −1)? — xⁿ⁺¹/(n+1) + C.∫xⁿ dx = xⁿ⁺¹/(n+1) + C. Increase the power by 1 and divide by the new power.
∫k·f(x) dx = ? — k·F(x) + C.Constants factor out of integrals: ∫k·f(x) dx = k·∫f(x) dx = k·F(x) + C.
To verify F(x) = (1/3)x³ is an antiderivative of x²: — Differentiate (1/3)x³ to get x².d/dx[(1/3)x³] = x². ✓ This confirms F(x) = (1/3)x³ is indeed an antiderivative.
∫(7x⁶) dx = ? — x⁷ + C.∫7x⁶ dx = 7·x⁷/7 + C = x⁷ + C.
The Fundamental Theorem of Calculus (FTC) Part 2 states: ∫_a^b f(x) dx = ? — F(b) − F(a).∫_a^b f(x) dx = F(b) − F(a), where F is any antiderivative of f. This connects integration and differentiation.
The definite integral ∫_a^b f(x) dx represents (when f ≥ 0): — The net signed area between f and the x-axis.For f ≥ 0, the definite integral gives the area under the curve. More generally, it gives the net signed area.
∫_a^a f(x) dx = ? — 0.When upper and lower limits are equal: F(a) − F(a) = 0. No area over zero width.
∫₂^5 k dx (k constant) = ? — Both B and C.F(x) = kx. F(5) − F(2) = 5k − 2k = 3k = k(5−2) = k·(b−a). Options B and C are the same.
FTC tells us that integration and differentiation are: — Inverse operations.The FTC reveals that differentiation and integration are inverse operations — a profound unifying result of calculus.
U-substitution is the integration counterpart of: — Chain rule.U-substitution reverses the chain rule of differentiation.
For ∫2x(x²+1)³ dx, choose u = ? — x²+1.Choose u = x² + 1 (the inner function). Then du = 2x dx, which is already present.
With u = x² + 1, du = 2x dx, the integral ∫2x(x²+1)³ dx becomes: — ∫u³ du.2x dx = du. Integral = ∫u³ du = u⁴/4 + C = (x²+1)⁴/4 + C.
∫₀^1 (x − x²) dx. First find intersections of y=x and y=x²: — x=0 and x=1.Set x = x²: x² − x = 0 → x(x−1) = 0 → x = 0 or x = 1.
Integration by parts formula: ∫u dv = ? — uv − ∫v du.∫u dv = uv − ∫v du. This is derived from the product rule: d(uv) = u dv + v du.
LIATE rule helps choose u. The first letter "L" stands for: — Logarithms.LIATE = Logarithms, Inverse trig, Algebraic, Trigonometric, Exponential. Higher priority items make better choices for u.
The "A" in LIATE stands for: — Algebraic (polynomials).A = Algebraic functions (polynomials like x, x², x³). These are often chosen as u when combined with exponentials or trig.
After applying IBP to ∫x eˣ dx: u=x, v=eˣ, we get xeˣ − ∫eˣ dx. The remaining integral ∫eˣ dx = ? — eˣ.∫eˣ dx = eˣ + C. So the full result is xeˣ − eˣ + C = eˣ(x−1) + C.