Integrate with ln?

Calculus Level 2

∫ 0 ln ⁡ e 3 x 3 − 2 x 2 6 d x \large \int_0^{\ln e} \dfrac{3x^3-2x^2}6\, dx

If the integral above can be expressed as a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 73.

Ashish Menon by Ashish Menon , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ashish Menon
May 20, 2016

ln ⁡ e = log ⁡ e e = 1 \ln{e} = {\log}_e^e = 1 .
So, ∫ 0 1 3 x 3 − 2 x 2 6 d x \int_{0}^{1} \dfrac{3x^3 - 2x^2}{6} \ dx = ∫ 0 1 1 2 x 3 − 1 3 x 2 d x = 1 2 ∫ 0 1 x 3 d x − 1 3 ∫ 0 1 x 2 d x = 1 2 × x 4 4 ∣ 0 1 − 1 3 × x 3 3 ∣ 0 1 = 1 2 × ( 1 4 4 − 0 4 4 ) − 1 3 ( 1 3 3 − 0 3 3 ) = 1 2 × 1 − 0 4 − 1 3 × 1 − 0 3 = 1 2 × 1 4 − 1 3 × 1 3 = 1 8 − 1 9 = 9 − 8 72 = 1 72 ∴ a + b = 1 + 72 = 73 = \int_{0}^{1} \dfrac{1}{2}x^3 - \dfrac{1}{3}x^2 \ dx\\ = \dfrac{1}{2} \int_{0}^{1} x^3 \ dx - \dfrac{1}{3} \int_{0}^{1} x^2 \ dx\\ = \dfrac{1}{2} × \dfrac{x^4}{4}{\Huge{\vert}}_{0}^{1} - \dfrac{1}{3} × \dfrac{x^3}{3}{\Huge{\vert}}_{0}^{1}\\ = \dfrac{1}{2} × \left( \dfrac{1^4}{4} - \dfrac{0^4}{4} \right) - \dfrac{1}{3} \left( \dfrac{1^3}{3} - \dfrac{0^3}{3} \right)\\ = \dfrac{1}{2} × \dfrac{1-0}{4} - \dfrac{1}{3} × \dfrac{1-0}{3}\\ = \dfrac{1}{2} × \dfrac{1}{4} - \dfrac{1}{3} × \dfrac{1}{3}\\ = \dfrac{1}{8} - \dfrac{1}{9}\\ = \dfrac{9 - 8}{72}\\ = \dfrac{1}{72}\\ \\ \therefore a + b = 1 + 72\\ = \color{#69047E}{\boxed{73}}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Same way..But thought ln e = 1 at last..:D

Sagar Shah - 5 years ago

(1/sin)/X|

Sujit Verma Don - 5 years ago

Log in to reply

1/sin x ????

Ashish Menon - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...