Calculus #1

Calculus Level 3

F ( x ) = ∫ x 2 x 3 ln ⁡ t d t \large F(x) = \int^{x^3}_{x^2} \ln t \ dt

F ( x ) F(x) is defined as above for x > 0 x>0 . If F ′ ( 2 ) = a ln ⁡ b F'(2) = a \ln b , where a a and b b are positive integers with b b being a prime, find a + b a+b .

Note: F ′ ( x ) F'(x) denotes the first derivative of F ( x ) F(x) with respect to x x .


The answer is 30.

Md Zuhair by Md Zuhair , 20, India

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2 solutions

Chew-Seong Cheong
Mar 29, 2017

F ( x ) = ∫ x 2 x 3 ln ⁡ t d t = t ln ⁡ t − t ∣ x 2 x 3 = 3 x 3 ln ⁡ x − x 3 − 2 x 2 ln ⁡ x + x 2 = ( 3 x 3 − 2 x 2 ) ln ⁡ x − x 3 + x 2 F ′ ( x ) = ( 9 x 2 − 4 x ) ln ⁡ x + ( 3 x 3 − 2 x 2 ) ⋅ 1 x − 3 x 2 + 2 x = ( 9 x 2 − 4 x ) ln ⁡ x + 3 x 2 − 2 x − 3 x 2 + 2 x = ( 9 x 2 − 4 x ) ln ⁡ 2 F ′ ( 2 ) = ( 36 − 8 ) ln ⁡ 2 = 28 ln ⁡ 2 \begin{aligned} F(x) & = \int_{x^2}^{x^3} \ln t \ dt \\ & = t \ln t - t \ \bigg|_{x^2}^{x^3} \\ & = 3x^3 \ln x - x^3 - 2 x^2 \ln x + x^2 \\ & = \left(3x^3 - 2 x^2 \right) \ln x - x^3 + x^2 \\ F'(x) & = \left(9x^2 - 4x \right) \ln x + \left(3x^3 - 2 x^2 \right) \cdot \frac 1x - 3x^2 + 2x \\ & = \left(9x^2 - 4x \right) \ln x + 3x^2 - 2x - 3x^2 + 2x \\ & = \left(9x^2 - 4x \right) \ln 2 \\ F'(2) & = \left(36 - 8 \right) \ln 2 \\ & = 28\ln 2 \end{aligned}

⟹ a + b = 28 + 2 = 30 \implies a+b = 28+2 = \boxed{30}

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Sir I think in the second line, it should be t ( l n t − 1 ) t(lnt-1)

Rahil Sehgal - 4 years, 2 months ago

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Yes, you are right.

Chew-Seong Cheong - 4 years, 2 months ago

Md, again ln is a function. You should add a backslash \ ln t ln ⁡ t \ln t . Note that ln is not italic and there is a space between ln and t, while ln t l n t ln t is italic and no space between ln and t. The answer should be 28 ln ⁡ 2 + 8 28\ln 2 + 8 . Please be care on the answer. Luckily for this problem we can solve it with a ln ⁡ b + b 3 a\ln b + b^3 .

Chew-Seong Cheong - 4 years, 2 months ago

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Sorry, the answer should be a ln ⁡ b a \ln b as before, I have changed it.

Chew-Seong Cheong - 4 years, 2 months ago

Thanks sir. You helped me a lot in learning up latex.

Md Zuhair - 4 years, 2 months ago

Sir in 2nd last line it would be 28ln2

Md Zuhair - 4 years, 2 months ago

I used the Newton Leibnitz's Formula

Ashutosh Kumar - 3 years, 12 months ago

Sir @Chew-Seong Cheong , which latex editor do you use for your solutions?

Sarthak Chaturvedi - 3 years, 11 months ago

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You can just enter the codes in the pages of this website. You can see the codes by clicking the pull-down menu ⋯ \cdots at the right bottom corner of the problem and select "Toggle LaTex". Or placing your mouse cursor on the formulas.

Chew-Seong Cheong - 3 years, 11 months ago

Sir in 2nd last line it would be 28ln2

Md Zuhair - 4 years, 2 months ago

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28ln2 is same as 28lnx as we are putting the value of x=2...

Rahil Sehgal - 4 years, 2 months ago

Thanks again.

Chew-Seong Cheong - 4 years, 2 months ago
Atul Kumar Ashish
Mar 29, 2017

We can use fundamental theorem of calculus to calculate F ′ ( 2 ) F'(2) without having to evaluate the integral. F ( x ) = ∫ x 2 x 3 ln ⁡ t d t = ∫ 0 x 3 ln ⁡ t d t − ∫ 0 x 2 ln ⁡ t d t \begin{aligned} F(x)&=\int_{x^2}^{x^3}\ln t\,dt\\&=\int_{0}^{x^3}\ln t\,dt-\int_{0}^{x^2}\ln t\,dt\end{aligned} Using chain rule, F ′ ( x ) = 3 x 2 ln ⁡ x 3 − 2 x ln ⁡ x 2 = 9 x 2 ln ⁡ x − 4 x ln ⁡ x = ( 9 x 2 − 4 x ) ln ⁡ x \begin{aligned} F'(x)&=3x^2\ln x^3-2x\ln x^2\\&=9x^2\ln x-4x \ln x\\&=(9x^2-4x)\ln x\end{aligned} Therefore, F ′ ( 2 ) = 28 ln ⁡ 2 a + b = 30 \begin{aligned}&F'(2)=28\ln 2\\&a+b=\boxed{30}\end{aligned}

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