trivial stuff

Calculus Level 3

∫ 0 2 ( 3 x 3 + x 2 ) 2 x + 1 d x = a 5 b \int_0^2 \left(3x^3 + x^{2}\right)\sqrt{2x+1}\ dx = \frac {a\sqrt 5}b

The equation above holds true for positive integers a a and b b . Find a + b a+b .


The answer is 43.

Sam Sam by Sam sam , 19, Finland

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

Consider the following:

y = ( 2 x 3 + x 2 ) 3 2 d y d x = 3 2 ( 2 x 3 + x 2 ) 1 2 ( 6 x 2 + 2 x ) = 3 ( 3 x 3 + x 2 ) 2 x + 1 \begin{aligned} y & = (2x^3+x^2)^\frac 32 \\ \frac {dy}{dx} & = \frac 32 (2x^3+x^2)^\frac 12 (6x^2+2x) = 3(3x^3+x^2)\sqrt{2x+1} \end{aligned}

⟹ I = ∫ 0 2 ( 3 x 3 + x 2 ) 2 x + 1 d x = 1 3 ∫ 0 2 d y d x d x = 1 3 ∫ 0 2 d y = y 3 = ( 2 x 3 + x 2 ) 3 2 3 ∣ 0 2 = 40 5 3 \begin{aligned} \implies I & = \int_0^2 (3x^3+x^2)\sqrt{2x+1}\ dx = \frac 13 \int_0^2 \frac {dy}{dx} dx = \frac 13 \int_0^2 dy \\ & = \frac y3 = \frac {(2x^3+x^2)^\frac 32}3 \ \bigg|_0^2 = \frac {40\sqrt 5}3 \end{aligned}

Therefore a + b = 40 + 3 = 43 a+b = 40 + 3 = \boxed {43} .

1 Helpful 0 Interesting 2 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...