Given F(x)=\int_{2x^{2}}^{x^{3}}\sqrt{2+\sqrt[3]{t}}dt, find the tangent line to F(x) at x = 2....

Question:

1. Given{eq}F(x)=\int_{2x^{2}}^{x^{3}}\sqrt{2+\sqrt[3]{t}}dt {/eq}, find the tangent line to F(x) at x = 2.

2. For {eq}x\geq 0 {/eq}, find {eq}F(x)=\int_{0}^{x} 6\left | t^{2}-t \right |dt. {/eq}

Fundamental Theorem of Calculus

  • If {eq}F(x) = \int_a^x f(t)dt {/eq}, then by the Fundamental Theorem of Calculus, {eq}F'(x) = f(x) {/eq}.
  • If {eq}F(x) = \int_a^{g(x)} f(t)dt {/eq}, then by the General Fundamental Theorem of Calculus, {eq}F'(x) = f(g(x))g'(x) {/eq}.

Below are some important integral properties:

  • {eq}\int_a^a f(x) dx = 0 {/eq}
  • {eq}\int_a^b f(x)dx = -\int_b^a f(x) dx {/eq}
  • {eq}\int_a^b f(x) dx = \int_a^c f(x)dx + \int_c^b f(x)dx {/eq}

Answer and Explanation:

(1) Evaluate {eq}F(2) {/eq}.

{eq}\begin{align} F(2) = \int_{8}^{8}\sqrt{2+\sqrt[3]{t}}dt = 0 \end{align} {/eq}

Rewrite {eq}F(x) {/eq} using integral properties.

{eq}\begin{align} F(x)&=\int_{2x^{2}}^{x^{3}}\sqrt{2+\sqrt[3]{t}}dt \\ &=\int_{2x^{2}}^{0}\sqrt{2+\sqrt[3]{t}}dt +\int_{0}^{x^{3}}\sqrt{2+\sqrt[3]{t}}dt \\ &=-\int_{0}^{2x^{2}}\sqrt{2+\sqrt[3]{t}}dt +\int_{0}^{x^{3}}\sqrt{2+\sqrt[3]{t}}dt \end{align} {/eq}

Evaluate {eq}F'(x) {/eq} using the General Fundemantal Theoreom of Calculus.

{eq}\begin{align} F'(x)&=-\sqrt{2+\sqrt[3]{2x^2}}(4x) + \sqrt{2+\sqrt[3]{x^3}}(3x^2) \\ &=-4x\sqrt{2+\sqrt[3]{2x^2}} +3x^2\sqrt{2+x} \end{align} {/eq}

Evaluate {eq}F'(2) {/eq}.

{eq}\begin{align} F'(2) &=-8\sqrt{2+2} +12\sqrt{2+2} \\ &=-16+24 \\ &=8 \end{align} {/eq}

The tangent line passes through the point {eq}(2,0) {/eq} and has slope {eq}8 {/eq}. Use the point-slope formula to find the equation of the tangent line.

{eq}\begin{align} y-0&=8(x-2)\\ y &=8x-16 \end{align} {/eq}

(2) Solve the equation {eq}t^2-t = 0 {/eq}.

{eq}\begin{align} t^2-t&=0\\ t(t-1)&=0 \\ t = 0, 1 \end{align} {/eq}

Using the signs of the factors of {eq}t^2-t = t(t-1) {/eq} we obtain the following results.

  • If {eq}0 \leq t \leq 1 {/eq}, then {eq}t^2-t \leq 0 {/eq}, and so {eq}|t^2-t| = -(t^2-t) = t-t^2 {/eq}.
  • If {eq}t > 1 {/eq}, then {eq}t^2-t >0 {/eq}, and so {eq}|t^2-t| = t^2-t {/eq}.

Evaluate {eq}F(x) {/eq} for {eq}0 \leq x \leq 1 {/eq}.

{eq}\begin{align} F(x)&=\int_{0}^{x} 6\left | t^{2}-t \right |dt \\ &=\int_0^x 6(t-t^2)dt \\ &=6\left(\frac{t^2}{2} -\frac{t^3}{3}\right)\Big|_{0}^{x} \\ &=3t^2-2t^3 \Big|_{0}^{x}\\ &=3x^2-2x^3 \end{align} {/eq}

Evaluate {eq}F(x) {/eq} for {eq}x> 1 {/eq}.

{eq}\begin{align} F(x)&=\int_{0}^{x} 6\left | t^{2}-t \right |dt \\ &= \int_{0}^{1} 6\left | t^{2}-t \right |dt + \int_{1}^{x} 6\left | t^{2}-t \right |dt \\ &=3(1)^2-2(1)^3+ \int_{1}^{x} 6\left( t^{2}-t \right )dt \\ &=1 + 6\left(\frac{t^3}{3} -\frac{t^2}{2}\right)\Big|_{1}^{x} \\ &= 1+(2t^3-3t^2) \Big|_{1}^{x} \\ &=1 +[(2x^3-2x^2)-(2(1)^3-3(1)^2)]\\ &=1+[2x^3-2x^2+1]\\ &=2+2x^3-2x^2 \end{align} {/eq}

Write {eq}F(x) {/eq} as piece-wise function.

{eq}\begin{align} F(x) = \begin{cases} 3x^2-2x^3 & \text{if } 0 \leq x \leq 1 \\ 2+2x^3-2x^2 &\text{if } x >1 \end{cases} \end{align} {/eq}


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The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10
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