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  1. Problems
  2. /A function has the derivative F'(x) = 6x²/(x³ + 2) for x ≥ 0, and F(0) = 0…
MediumMathematicsHigh School (11–12)Functions and Calculus › IntegralsNumericalA-LevelAPBoard ExamIB

A function has the derivative F'(x) = 6x²/(x³ + 2) for x ≥ 0, and F(0) = 0…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A function has the derivative F′(x)=6x2x3+2F'(x) = \dfrac{6x^2}{x^3 + 2}F′(x)=x3+26x2​ for x≥0x \ge 0x≥0, and F(0)=0F(0) = 0F(0)=0. Use the substitution u=x3+2u = x^3 + 2u=x3+2. Find the constant of integration CCC in F(x)=2ln⁡(x3+2)+CF(x) = 2\ln(x^3 + 2) + CF(x)=2ln(x3+2)+C, and then F(2)F(2)F(2), correct to four decimal places.

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Work it out on paper, then type your number. The answer itself stays hidden.

Need a reminder? The key ideaShowHide

Put u=g(x)u = g(x)u=g(x) and du=g′(x) dxdu = g'(x)\,dxdu=g′(x)dx, so that ∫f(g(x)) g′(x) dx=∫f(u) du\int f(g(x))\,g'(x)\,dx = \int f(u)\,du∫f(g(x))g′(x)dx=∫f(u)du. Replace every xxx, integrate in uuu, then return to xxx.

∫f′(x)f(x) dx=ln⁡∣f(x)∣+C.\displaystyle \int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C.∫f(x)f′(x)​dx=ln∣f(x)∣+C.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Substitute. Let u=x3+2u = x^3 + 2u=x3+2. Then du=3x2 dxdu = 3x^2\,dxdu=3x2dx, so 6x2 dx=2 du6x^2\,dx = 2\,du6x2dx=2du. The integral is 2∫duu=2ln⁡∣u∣+C2\displaystyle\int \dfrac{du}{u} = 2\ln|u| + C2∫udu​=2ln∣u∣+C.

Step 2. The constant. Since x3+2>0x^3 + 2 > 0x3+2>0, F(x)=2ln⁡(x3+2)+CF(x) = 2\ln(x^3 + 2) + CF(x)=2ln(x3+2)+C. At x=0x = 0x=0: 0=2ln⁡2+C0 = 2\ln 2 + C0=2ln2+C, so C=−2ln⁡2=−1.3863C = -2\ln 2 = -1.3863C=−2ln2=−1.3863.

Step 3. Evaluate. F(2)=2ln⁡10−2ln⁡2=2ln⁡5=3.2189F(2) = 2\ln 10 - 2\ln 2 = 2\ln 5 = 3.2189F(2)=2ln10−2ln2=2ln5=3.2189.

Answer: C=−1.3863C = -1.3863C=−1.3863 and F(2)=3.2189F(2) = 3.2189F(2)=3.2189.

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