Putting the Two-Limit Formula to Work
The formula \(\dfrac{d}{dx}\displaystyle\int_{a(x)}^{b(x)}f(t)\,dt=f(b(x))\,b'(x)-f(a(x))\,a'(x)\) is easy to state and easy to misapply under pressure. This tutorial works through complete problems that are typical of exams: a tangent line, a closed-form simplification, and an optimization. In each, we compute the derivative with the formula and then confirm it by another route, because a result that has been checked twice can be trusted.
Worked Example 1: A Tangent Line
Let \(y=\displaystyle\int_x^{x^2}(2t+1)\,dt\). Find the equation of the tangent line at \(x=1\).
The point. At \(x=1\) both limits equal \(1\), so the integral is over an interval of length zero and \(y(1)=0\).
The slope. With \(f(t)=2t+1\), \(b=x^2\), \(a=x\):
At \(x=1\), \(y'(1)=3\). The tangent line is \(y=3(x-1)\).
Check by direct integration: the antiderivative of \(2t+1\) is \(t^2+t\), so \(y=(x^4+x^2)-(x^2+x)=x^4-x\). Then \(y(1)=0\) and \(y'=4x^3-1\), which equals \(3\) at \(x=1\). The two methods agree.
Worked Example 2: A Hidden Simplification
Let \(G(x)=\displaystyle\int_{1/x}^{x}\dfrac{dt}{t}\) for \(x>0\). Find \(G'(x)\) and simplify \(G\).
Here \(b=x\) with \(b'=1\), and \(a=1/x\) with \(a'=-1/x^2\). Then
Direct integration gives \(G(x)=\ln x-\ln(1/x)=\ln x+\ln x=2\ln x\), whose derivative is \(2/x\). The formula and the closed form match. Notice that the second term in the formula is added in effect, because the lower limit moves left when \(x\) increases and so contributes area.
Numerical check at \(x=2\): \(G'(2)=1\), and a numerical derivative of the integral gives 1.0.
Worked Example 3: Locating a Maximum
Let \(K(x)=\displaystyle\int_x^{x+1}e^{-t^2}\,dt\), the area under the bell curve over a window of width one. Where is the window placed to capture the most area?
Here \(b=x+1\), \(a=x\), and both have derivative \(1\). So
Setting \(K'=0\) requires \((x+1)^2=x^2\), which gives \(2x+1=0\), so \(x=-\tfrac12\). For \(x<-\tfrac12\), \((x+1)^2<x^2\), so the first exponent is less negative and \(K'>0\). For \(x>-\tfrac12\), \(K'<0\). Thus \(K\) has a maximum at \(x=-\tfrac12\), where the window \([-\tfrac12,\tfrac12]\) is centered at the peak.
Value and check: \(K(-\tfrac12)=\displaystyle\int_{-1/2}^{1/2}e^{-t^2}dt\approx0.922562\). At \(x=0.3\), the formula gives \(K'(0.3)=e^{-1.69}-e^{-0.09}=-0.729412\), matching a numerical derivative. The negative sign is correct because \(0.3>-\tfrac12\).
Summary of Problem Types
| Problem type | What to compute | Example result |
|---|---|---|
| Tangent line | y(x₀) and y'(x₀) from the formula | y = 3(x − 1) |
| Simplify first | Combine logarithm or algebra before differentiating | G = 2 ln x, G' = 2/x |
| Maximum or minimum | Solve f(b)b' = f(a)a' | x = −1/2 for the window |
| Verify | Integrate in closed form or differentiate numerically | Both agree at each test point |
Common Mistakes
Forgetting a chain factor on a limit like \(1/x\). In Example 2, using \(a'=1\) instead of \(a'=-1/x^2\) would give the wrong sign for the second term and produce \(0\) instead of \(2/x\).
Setting \(F'=0\) incorrectly. In Example 3, writing \(e^{-(x+1)^2}=0\) is not valid; the equation is \(e^{-(x+1)^2}=e^{-x^2}\), which is solved by comparing exponents.
Skipping the point on the curve. A tangent line requires both the value and the slope. In Example 1, the value is zero because the limits coincide at \(x=1\), a fact that is easy to overlook.
Differentiate First or Integrate First?
When the integrand has an elementary antiderivative, there are two valid routes: integrate and then differentiate, or apply the formula directly. Both give the same answer, and each is a check on the other. The formula route is shorter when the antiderivative is messy, and it is the only route when no elementary antiderivative exists, as with \(e^{-t^2}\) or \(\sin(t^2)\). The integrate-first route is shorter when the integrand is a simple polynomial, because the result is a polynomial that can be differentiated term by term and also analyzed for concavity and extrema.
Worked Example 4: Second Derivative and Concavity
Return to \(y=\displaystyle\int_x^{x^2}(2t+1)\,dt=x^4-x\). Find where the graph is concave up.
We have \(y'=4x^3-1\), so \(y''=12x^2\). This is nonnegative for all \(x\) and zero only at \(x=0\), so the graph is concave up everywhere except for a single point of zero curvature at the origin. The critical point is where \(4x^3=1\), that is \(x=4^{-1/3}\approx0.62996\), and since \(y''>0\) there, it is a minimum.
The same conclusion can be reached without integrating: the formula gives \(y'=4x^3-1\) directly, and differentiating that again is routine. The integral form shows why the derivative of the lower limit contributed the constant \(-1\).
Worked Example 5: A Logarithmic Integrand
Let \(A(x)=\displaystyle\int_x^{3x}\ln(1+t^2)\,dt\). Find \(A'(1)\).
The upper limit is \(3x\) with derivative \(3\), and the lower limit is \(x\) with derivative \(1\). So \(A'(x)=3\ln(1+9x^2)-\ln(1+x^2)\), and at \(x=1\),
Numerically, \(3\ln10-\ln2=6.21461\), and a numerical derivative of the integral at \(x=1\) gives 6.21461. The integrand has no convenient antiderivative to check against, so the numerical derivative is the independent check.
Reading the Result
After computing a derivative of this kind, ask whether its sign makes sense. In Example 5, the integrand \(\ln(1+t^2)\) increases for \(t>0\), and both limits move to the right as \(x\) grows, so the interval slides toward larger values and also lengthens. The area should increase quickly, and indeed \(A'(1)\approx6.215\) is positive and large. In Example 3, the window moves right across the peak of the bell curve, so the area first increases and then decreases, matching the sign change of \(K'\). Such sanity checks catch the most common errors, which are sign mistakes and missing chain factors.
One more caution concerns domains. All of the formulas above assume that \(f\) is continuous on an interval containing every value taken by both limits. In Example 2, the integrand \(1/t\) is continuous on \((0,\infty)\), and both \(x\) and \(1/x\) stay positive when \(x>0\), so the computation is legitimate. If the limits ever crossed zero, the integral would be undefined. Checking the domain takes seconds and prevents meaningless answers.
It also helps to remember that the formula is linear in the integrand: if \(f=f_1+f_2\), the derivative splits into the corresponding sum. Combined with the checklist, this lets you break a complicated integrand into pieces and differentiate each one separately, then add the results.
Where This Fits in the Course
This tutorial belongs to the module “The Fundamental Theorem of Calculus”. The tutorial just before it is “Variable-Limit Derivatives With Trigonometric Integrands,” and the one that follows is “The Missing Chain Factor Error.”
Practice Problems with Answers
Try each problem before reading the answer. Each answer below was verified by differentiation or by evaluating a difference quotient or integral numerically.
- Differentiate \(\int_{x}^{x^2}\dfrac{dt}{1+t^2}\).
Answer. \(\dfrac{2x}{1+x^4}-\dfrac1{1+x^2}\). - Find the tangent line to \(y=\int_{x}^{2x}(3t^2)\,dt\) at \(x=1\).
Answer. \(y=7x^3\), so \(y(1)=7\), \(y'(1)=21\), tangent \(y=7+21(x-1)\). - For \(G(x)=\int_{1/x}^{x}\dfrac{dt}{t}\), compute \(G'(5)\).
Answer. \(2/5=0.4\). - Where does \(\int_x^{x+2}e^{-t^2}dt\) have its maximum?
Answer. \(e^{-(x+2)^2}=e^{-x^2}\) gives \(x=-1\).
Check Your Understanding
Work these without looking back.
- Differentiate \(\int_{2x}^{x^2}\sqrt{1+t^4}\,dt\).
- Find \(\dfrac{d}{dx}\int_{-x}^{x}f(t)\,dt\) in terms of \(f\).
- Simplify \(\int_{x}^{x^2}2t\,dt\) first, then differentiate. Compare with the formula.
- Explain why \(y(1)=0\) in Example 1 without computing any antiderivative.
- Find the critical points of \(\int_x^{x+1}\sin(t^2)\,dt\) in terms of an equation.
