Integrals
The fundamental theorem of calculus
Differentiation and integration are linked: differentiating accumulated area recovers the local integrand, and primitives evaluate definite integrals.
1. Differentiate accumulation
1 / 3A(x)=\int_0^x t^2dt=\frac{x^3}{3},\quad A\prime(x)=x^2 Let A(x)=∫₀ˣt²dt=x³/3. A tiny change in x adds a strip with height x², so A′(x)=x². The accumulated area and its rate are different values.
- A(b)
- 0.333333
- A′(b)
- 1
Shading shows the region between the bounds. With bounds in increasing order, regions below the axis contribute negatively; reversing the bounds reverses the integral’s sign. Geometric area stays nonnegative.
2. Evaluate with a primitive
2 / 3\int_0^b e^t\,dt=e^b-1 Since (eᵗ)′=eᵗ, ∫₀ᵇeᵗdt=eᵇ−1. This evaluates the limiting sum exactly without drawing infinitely many rectangles.
- eᵇ
- 2.718282
- ∫
- 1.718282
Shading shows the region between the bounds. With bounds in increasing order, regions below the axis contribute negatively; reversing the bounds reverses the integral’s sign. Geometric area stays nonnegative.
3. A moving upper limit
3 / 3A(x)=\int_0^{x^2}\cos t\,dt,\quad A\prime(x)=2x\cos(x^2) For A(x)=∫₀^(x²)cos(t)dt=sin(x²), differentiate the accumulation at the upper limit and multiply by its derivative: A′(x)=cos(x²)·2x.
- upper=x²
- 1
- A(a)
- 0.841471
- A′(a)
- 1.080605
The solid curve is f; the dashed line is its tangent at a. Read f(a) as height and f′(a) as slope. Questions use the input stated in their text.