Question

Prove that \(\frac{d}{d x}\left[\int_{u(x)}^{v(x)} f(t) d t\right]=f(v(x)) v^{\prime}(x)-f(u(x)) u^{\prime}(x)\).

Short answer

The proof of the given exercise is based on the application of the First Fundamental Theorem of Calculus together with the Chain Rule. First, the definite integral is rewritten using the Fundamental Theorem of Calculus. Then, the Chain Rule is applied to derive the desired result.

Step-by-step solution

Apply the Fundamental Theorem of Calculus Part 1

Take the definite integral \( \int_{u(x)}^{v(x)} f(t) dt \), where \( u(x) \) and \( v(x) \) are the limits of the integral. By applying the First Fundamental Theorem of Calculus, we can rewrite the integral as \( F(v(x)) - F(u(x)) \), where F is an antiderivative of f.

Differentiate Using the Chain Rule

Now, differentiate \( F(v(x)) - F(u(x)) \) with respect to \( x \). Using the chain rule yields \( f(v(x))v'(x) - f(u(x))u'(x) \), where \( f(v(x)) \) and \( f(u(x)) \) are the derivatives of \( F \) at \( v(x) \) and \( u(x) \), respectively. The terms \( v'(x) \) and \( u'(x) \) are the derivatives of the functions \( v(x) \) and \( u(x) \), respectively.

Conclusion

We have now derived the result \(\frac{d}{d x}\left[\int_{u(x)}^{v(x)} f(t) d t\right]=f(v(x)) v^{\prime}(x)-f(u(x)) u^{\prime}(x)\), which is the formula for differentiation under the integral sign, often attributed to Leibniz.