Second Fundamental Theorem of Calculus – Questions with Answers

This page presents calculus questions with detailed solutions based on the Second Fundamental Theorem of Calculus.

Theorem

The second fundamental theorem of calculus states that if \( f \) is continuous on an interval \( I \) containing \( a \), and if

\[ F(x) = \int_a^x f(t)\,dt \]

then

\[ F'(x) = f(x) \]

for all \( x \in I \).

Approximate \( F'(\pi/2) \) to three decimal places if

\[ F(x) = \int_3^x \sin(t^2)\,dt \]

Since \( \sin(t^2) \) is continuous for all real \( t \), the Second Fundamental Theorem applies: \[ F'(x) = \sin(x^2) \]
Evaluating at \( x = \pi/2 \): \[ F'(\pi/2) = \sin\!\left(\left(\frac{\pi}{2}\right)^2\right) \approx 0.624 \]

Let

\[ F(x) = \int_0^x \frac{5}{3 + 2e^t}\,dt \]

a) Find \( F'(0) \).
b) Show that \( F(1) < F(4) \).

Since \( \frac{5}{3 + 2e^t} \) is continuous, the theorem gives: \[ F'(x) = \frac{5}{3 + 2e^x} \]
Evaluating at \( x = 0 \): \[ F'(0) = \frac{5}{3 + 2e^0} = \frac{5}{5} = 1 \]
Because \( F'(x) > 0 \) for all \( x \), the function \( F \) is increasing. Since \( 4 > 1 \), it follows that \[ F(1) < F(4) \]

Let

\[ F(x) = \int_{-1}^{x^2} \frac{1}{1+t^2}\,dt \]

Find \( F'(x) \).

Let \( u = x^2 \). Then \[ F(u) = \int_{-1}^{u} \frac{1}{1+t^2}\,dt \]
By the Second Fundamental Theorem: \[ \frac{dF}{du} = \frac{1}{1+u^2} \]
Applying the chain rule: \[ F'(x) = \frac{dF}{du}\cdot\frac{du}{dx} = \frac{1}{1+x^4}\cdot 2x = \frac{2x}{1+x^4} \]

Let

\[ F(x) = \int_{u(x)}^{v(x)} f(t)\,dt \]

where \( f \) is continuous and \( u \), \( v \) are differentiable functions of \( x \). Express \( F'(x) \).

Rewrite the integral using a fixed constant \( a \): \[ F(x) = -\int_a^{u(x)} f(t)\,dt + \int_a^{v(x)} f(t)\,dt \]
By the Second Fundamental Theorem: \[ \frac{d}{du}\!\left(-\int_a^u f(t)\,dt\right) = -f(u), \quad \frac{d}{dv}\!\left(\int_a^v f(t)\,dt\right) = f(v) \]
Applying the chain rule: \[ F'(x) = v'(x)f(v(x)) - u'(x)f(u(x)) \]