How to Use the Fundamental Theorem of Calculus: A Complete Guide

The Fundamental Theorem of Calculus connects differentiation and integration, showing they are inverse processes. It allows us to compute definite integrals without using Riemann sums, provided we can find an antiderivative. This theorem is essential for solving problems involving area, motion, and accumulation in fields like physics and engineering. It has two main parts, each revealing a different aspect of how derivatives and integrals relate.

Understanding the Two Parts

The first part states that if $ f $ is continuous on $[a, b]$, then the function $ F(x) = \int_a^x f(t) , dt $ is an antiderivative of $ f $. That is, $ F’(x) = f(x) $. This guarantees that every continuous function has an antiderivative, even if we can’t express it in simple terms Wikipedia — Fundamental theorem of calculus.

The second part is used more often in calculations. It says that if $ F $ is any antiderivative of $ f $, then $ \int_a^b f(x) , dx = F(b) - F(a) $. This means we can evaluate definite integrals by finding an antiderivative and subtracting its values at the endpoints. Without this, we’d have to rely on approximations for every integral.

Together, these parts show that integrating a function and then differentiating brings you back to the original function, and vice versa—differentiation undoes integration.

Applying the First Part

The first part is useful when working with functions defined as integrals. For example, if $ F(x) = \int_1^x \sin(t^2) , dt $, we cannot write a simple formula for $ F(x) $, but we know immediately that $ F’(x) = \sin(x^2) $, since the integrand is continuous.

This rule extends when the upper limit is a function of $ x $. Suppose $ G(x) = \int_1^{x^2} e^{t^3} , dt $. Let $ u = x^2 $, so $ G(x) = \int_1^u e^{t^3} , dt $. Then by the chain rule, $ \frac{dG}{dx} = \frac{dG}{du} \cdot \frac{du}{dx} = e^{u^3} \cdot 2x = 2x e^{x^6} $.

If the lower limit is the variable, use the property $ \int_a^b f(t),dt = -\int_b^a f(t),dt $. For $ H(x) = \int_x^3 \ln(t+1) , dt $, rewrite it as $ -\int_3^x \ln(t+1) , dt $, so $ H’(x) = -\ln(x+1) $.

When both limits depend on $ x $, differentiate each part separately. For $ K(x) = \int_x^{x^2} \cos(t) , dt $, split it at a constant $ c $:
$ K(x) = \int_x^c \cos(t),dt + \int_c^{x^2} \cos(t),dt = -\int_c^x \cos(t),dt + \int_c^{x^2} \cos(t),dt $.
Then $ K’(x) = -\cos(x) + \cos(x^2) \cdot 2x = 2x\cos(x^2) - \cos(x) $ Paul’s Online Math Notes.

Using the Second Part to Evaluate Integrals

The second part is the main tool for computing definite integrals. To use it, find an antiderivative of the integrand, evaluate it at the upper and lower limits, and subtract.

For example, compute $ \int_1^4 3x^2 , dx $. An antiderivative of $ 3x^2 $ is $ x^3 $. Then:
$ \int_1^4 3x^2 , dx = \left[ x^3 \right]_1^4 = 4^3 - 1^3 = 64 - 1 = 63 $.

This method works for many functions—polynomials, trigonometric, exponential—as long as an antiderivative exists. For $ \int_0^{\pi} \sin(x) , dx $, the antiderivative is $ -\cos(x) $:
$ \left[ -\cos(x) \right]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 $.
This gives the net area under one hump of the sine curve.

Some functions, like $ e^{-x^2} $ or $ \frac{\sin(x)}{x} $, do not have elementary antiderivatives. The second part cannot be applied directly in these cases, and numerical methods are needed. However, the first part still applies: $ \int_0^x e^{-t^2} dt $ defines a function (related to the error function) whose derivative is $ e^{-x^2} $ Khan Academy.

Common Mistakes

A key requirement for the second part is continuity on the interval $[a, b]$. If the function has a discontinuity, the theorem does not apply. For example, $ \int_{-1}^1 \frac{1}{x^2} , dx $ cannot be evaluated using $ -\frac{1}{x} $ because the function is undefined at $ x = 0 $. The integral diverges, but blindly applying the theorem gives $ -1 - (-(-1)) = -2 $, which is wrong.

Another frequent error is mishandling the chain rule with variable limits. Always check whether the upper or lower limit depends on $ x $. If the variable is in the lower limit, include a negative sign. If both limits vary, treat each term separately and combine the derivatives.

Also, do not include the constant of integration ($ +C $) when evaluating definite integrals. It cancels out in the subtraction $ F(b) - F(a) $. Whether you use $ x^2 $ or $ x^2 + 7 $ as an antiderivative of $ 2x $, the result is the same: $ (b^2 + 7) - (a^2 + 7) = b^2 - a^2 $. So the choice of antiderivative does not affect the value of the definite integral MIT OpenCourseWare.

Key Takeaways

• The Fundamental Theorem of Calculus shows that differentiation and integration are inverse operations.
• The first part lets you differentiate functions defined by integrals, especially when the upper limit is a function of $ x $.
• The second part allows evaluation of definite integrals using antiderivatives: $ \int_a^b f(x),dx = F(b) - F(a) $.
• Always verify continuity on the interval before applying the second part, and apply the chain rule carefully when limits are functions of $ x $.