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Definite Integral Evaluator

Math Calculus • Integrals

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2. Definite Integral Evaluator
Evaluates \(\int_a^b f(x)\,dx\) using an antiderivative when possible (FTC), with a numerical fallback and a shaded-area plot.
Inputs
Supported: + − * / ^, parentheses, variable x, constants pi, e, sin cos tan, ln log(base 10), sqrt abs exp. Implicit multiplication is allowed: 2x, (x+1)(x-1), 2sin(x), xsin(x). Trig powers like cos^2(2x) are supported.
You can use pi, e, e.g. \(a=0\).
Example: \(b=\pi\).

If symbolic rules don’t match, use numeric.
LaTeX is best for readability.
Used in adaptive Simpson.

Plots use \(x\) on the horizontal axis.
Plots \(x\in[c-w,c+w]\). Use Auto fit if scales explode.
Does not change \(F'(x)\).
Click to auto-fill and compute.
Ready
Graph
Drag to pan • wheel/pinch to zoom • curves: \(f(x)\), \(F(x)+C\), \(F'(x)\). Shaded region shows the signed area on \([a,b]\).
\(f(x)\) \(F(x)+C\) \(F'(x)\)
x: 0, y: 0, zoom(px/unit): 60
Tip: If the plot scales explode, zoom out or use Auto fit.
Result
Enter \(f(x)\), bounds \(a,b\), and click Evaluate.

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Frequently Asked Questions

What does a definite integral evaluator calculate?

It calculates the accumulated signed area under y = f(x) from x = a to x = b. The result is a single number representing the net area above the x-axis minus the area below it.

How does the Fundamental Theorem of Calculus compute a definite integral?

If an antiderivative F(x) is known for f(x), then the integral from a to b of f(x) dx equals F(b) - F(a). This turns the area problem into evaluating an antiderivative at the bounds.

When will the calculator use numerical integration instead of symbolic FTC?

If the integrand does not match the calculator's symbolic antiderivative rules or has no simple elementary antiderivative, it can fall back to adaptive Simpson integration. Numerical methods approximate the area to within the chosen tolerance.

What does the numeric tolerance control in Simpson integration?

It sets the error target for the adaptive Simpson algorithm. Smaller tolerances generally increase accuracy but may require more subdivisions and computation.

Why does the constant C not change the definite integral value?

Adding a constant shifts an antiderivative up or down, but it cancels in F(b) - F(a). Both endpoints include the same added C, so the difference stays the same.