Consider the function

MATHEMATICS · Single correct · CBSE · Mathematics · Integrals · Integrals

Problem

Consider the function F(x)=∫1xt2+1 dtF(x)=\displaystyle \int_{1}^{x} \sqrt{t^{2}+1}\,dt. Which statement best justifies the conclusion that F′(x)=x2+1F'(x)=\sqrt{x^{2}+1}? (a) Because ddx∫1xf(t) dt=f′(x)\displaystyle \frac{d}{dx}\int_{1}^{x}f(t)\,dt = f'(x), differentiating the integral gives F′(x)=x2+1F'(x)=\sqrt{x^{2}+1}. (Incorrect use of FTC) (b) The Fundamental Theorem of Calculus says ddx∫1xf(t) dt=f(x)−f(1)\displaystyle \frac{d}{dx}\int_{1}^{x}f(t)\,dt = f(x)-f(1); substituting yields F′(x)=x2+1F'(x)=\sqrt{x^{2}+1}. (Misremembered formula) (c) The Fundamental Theorem of Calculus states that if ff is continuous on [1,x][1,x], then ddx∫1xf(t) dt=f(x)\displaystyle \frac{d}{dx}\int_{1}^{x} f(t)\,dt = f(x). Because t2+1\sqrt{t^{2}+1} is continuous, F′(x)=x2+1F'(x)=\sqrt{x^{2}+1}. (Correct) (d) Using the chain rule we first differentiate t2+1\sqrt{t^{2}+1} to obtain tt2+1\dfrac{t}{\sqrt{t^{2}+1}} and then substitute t=xt=x, so F′(x)=xx2+1F'(x)=\dfrac{x}{\sqrt{x^{2}+1}}. (Differentiated the integrand instead of the integral)

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