Science:Math Exam Resources/Courses/MATH101/April 2017/Question 06 (a)

MATH101 April 2017

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Question 06 (a)
Find the slope of the tangent line to the curve $y=\int _{1-x}^{1+2x}e^{t^{2}}dt$ at the point $(0,0)$. Simplify your answer completely.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Try to use the derivative formula for integrals (Fundamental theorem of calculus), and recall that the slope of tangent line is just the derivative at this point.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. According to the fundamental theorem of calculus, if we have following expression $$F(x)=\int _{a}^{x}h(t)\ dt$$ where ${\textstyle h(t)}$ is continuous real valued function on some interval ${\textstyle I}$, then its derivative is just ${\textstyle F'(x)=h(x)}$. Now if we have expression like $$F(x)=\int _{a}^{f(x)}h(t)\ dt$$ what is the derivative of ${\textstyle F(x)}$? we think above equation as follows $$G(y)=\int _{a}^{y}h(t)\ dt$$ $$F(x)=G(f(x))$$ Then by chain rule and fundamental theorem of calculus, we know that $$F'(x)=G'(f(x))\cdot f'(x)=h(f(x))\cdot f'(x)$$ . Now if we have expression like $$F(x)=\int _{g(x)}^{f(x)}h(t)\ dt$$ we can write it as a combination $$F(x)=\int _{0}^{f(x)}h(t)\ dt+\int _{g(x)}^{0}h(t)\ dt=\int _{0}^{f(x)}h(t)\ dt-\int _{0}^{g(x)}h(t)\ dt$$ Therefore, using above formula we have $$F'(x)=h(f(x))\cdot f'(x)-h(g(x))g'(x)$$ This formula can be seen in any calculus book which is very important for calculating the derivatives of integrals. Now we apply it in this question, in this case $$y=F(x),\ \ g(x)=1-x,\ \ f(x)=1+2x,\ \ h(t)=e^{t^{2}}$$ So $$F'(x)=2e^{(1+2x)^{2}}+e^{(1-x)^{2}}$$ We know that the slope of tangent line at point ${\textstyle (0,0)}$ is just the derivative of ${\textstyle y=F(x)}$ at this point. Especially, when ${\textstyle x=0,}$ the derivative is ${\textstyle \color {blue}F'(0)=3e}$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The slope of the tangent line containing $(0,0)$ is $m={\frac {dy}{dx}}{\bigg |}_{x=0}$. To find $m$, consider the function defined by $$g(x)=\int _{0}^{x}e^{t^{2}}\,dt$$ Then we can re-write $y=\int _{1-x}^{1+2x}e^{t^{2}}\,dt$ as $y=g(1+2x)-g(1-x)$. By the Fundamental Theorem of Calculus, $${\frac {d}{dx}}g(x)={\frac {d}{dx}}\int _{0}^{x}e^{t^{2}}\,dt=e^{x^{2}}$$ So, by the Chain Rule, $${\frac {dy}{dx}}={\frac {d}{dx}}(1+2x)\cdot e^{(1+2x)^{2}}-{\frac {d}{dx}}(1-x)\cdot e^{(1-x)^{2}}=2e^{(1+2x)^{2}}+e^{(1-x)^{2}}$$ Hence, plugging $x=0$, we get $$m={\frac {dy}{dx}}{\bigg |}_{x=0}=2e^{(1+2\times 0)^{2}}+e^{(1-0)^{2}}=3e$$ So, the slope of the tangent line is $\color {blue}3e$