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Definite Integrals and the Fundamental Theorem of Calculus

If F is an antiderivative of f on [a,b], then ∫_a^b f(x) dx=F(b)-F(a). This result is the evaluation form of the Fundamental Theorem of Calculus.

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Student-friendly explanation

A definite integral gives a number, not a family of functions. The limits fix the interval, so no arbitrary constant is written in the final value. The function should be integrable on the interval, and the antiderivative used must be valid throughout the interval.

How to write this in exams

  1. 1

    Start with the exact idea

    If F is an antiderivative of f on [a,b], then ∫_a^b f(x) dx=F(b)-F(a). This result is the evaluation form of the Fundamental Theorem of Calculus.

  2. 2

    Then show how to use it

    Check the limits and integrand. Find a valid antiderivative. Write [F(x)] from a to b. Substitute the upper limit first and lower limit second. Simplify F(b)-F(a). Do not add +C.

  3. 3

    Add one concrete example

    ∫_0^2 x^2 dx=[x^3/3]_0^2=8/3-0=8/3.

  4. 4

    Avoid this incomplete answer

    For ∫_0^π sinx dx, writing cosπ-cos0=-2 because the antiderivative of sinx was taken as cosx instead of -cosx.

Definition

If F is an antiderivative of f on [a,b], then ∫_a^b f(x) dx=F(b)-F(a). This result is the evaluation form of the Fundamental Theorem of Calculus.

Example

∫_0^2 x^2 dx=[x^3/3]_0^2=8/3-0=8/3.

Rule to remember

FTC evaluation rule: ∫_a^b f(x) dx=F(b)-F(a), where F'(x)=f(x) and f is integrable on [a,b]. No +C appears in the final definite integral value because the constant cancels.

Memory hook

Upper value minus lower value; the constant cancels.

Examples and method

Worked example

Evaluate ∫_0^π sinx dx. Antiderivative of sinx is -cosx. Therefore ∫_0^π sinx dx=[-cosx]_0^π=-cosπ-(-cos0)=1-(-1)=2. Final value: 2.

Method to apply

Check the limits and integrand. Find a valid antiderivative. Write [F(x)] from a to b. Substitute the upper limit first and lower limit second. Simplify F(b)-F(a). Do not add +C.

Diagram support

A graph may help interpret signed area, but routine FTC evaluation does not require a diagram. If used, label x=a, x=b, y=f(x), and signed regions above or below the x-axis.

How CBSE asks it

Asked as direct definite integral evaluation, as part of area/accumulation interpretation, or after using substitution, identities, partial fractions, or by-parts to find the antiderivative.

Avoid common mistakes

Common confusion

Students sometimes add +C in a definite integral or substitute limits in the wrong order as F(a)-F(b).

Common wrong answer

For ∫_0^π sinx dx, writing cosπ-cos0=-2 because the antiderivative of sinx was taken as cosx instead of -cosx.

Exam tip

Always write the antiderivative in bracket notation before substituting limits; it reduces sign and order errors.

Quick check

Evaluate ∫_1^3 2x dx.

[x^2]_1^3=9-1=8.

Answer writing and exam use

1-mark answer

If F is an antiderivative of f on [a,b], then ∫_a^b f(x) dx=F(b)-F(a). This result is the evaluation form of the Fundamental Theorem of Calculus.

2-mark answer

If F is an antiderivative of f on [a,b], then ∫_a^b f(x) dx=F(b)-F(a). This result is the evaluation form of the Fundamental Theorem of Calculus. FTC evaluation rule: ∫_a^b f(x) dx=F(b)-F(a), where F'(x)=f(x) and f is integrable on [a,b]. No +C appears in the final definite integral value because the constant cancels. ∫_0^2 x^2 dx=[x^3/3]_0^2=8/3-0=8/3.

3-mark answer

A definite integral gives a number, not a family of functions. The limits fix the interval, so no arbitrary constant is written in the final value. The function should be integrable on the interval, and the antiderivative used must be valid throughout the interval. FTC evaluation rule: ∫_a^b f(x) dx=F(b)-F(a), where F'(x)=f(x) and f is integrable on [a,b]. No +C appears in the final definite integral value because the constant cancels. Evaluate ∫_0^π sinx dx. Antiderivative of sinx is -cosx. Therefore ∫_0^π sinx dx=[-cosx]_0^π=-cosπ-(-cos0)=1-(-1)=2. Final value: 2. Asked as direct definite integral evaluation, as part of area/accumulation interpretation, or after using substitution, identities, partial fractions, or by-parts to find the antiderivative. For ∫_0^π sinx dx, writing cosπ-cos0=-2 because the antiderivative of sinx was taken as cosx instead of -cosx.
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