Fundamental theorem of integral calculus

Calculus Level 3

$\large y=\int_{-1}^{x}\frac{t^2}{t^2+4}\,dt-\int_{3}^{x}\frac{t^2}{t^2+4}\,dt$ If $y=f(x)$ , then find $f'(x)$ .

The answer is 0.

1 solution

Chew-Seong Cheong
Mar 22, 2017

$\begin{aligned} y & = \int_{-1}^x \frac {t^2}{t^2+4} dt - \int_{3}^x \frac {t^2}{t^2+4} dt \\ & = \int_{-1}^3 \frac {t^2}{t^2+4} dt = \color{#3D99F6}c \quad \small \text{where }c \text{ is a constant.} \end{aligned}$

$\begin{aligned} \implies f(x) & = c \\ f'(x) & = \boxed{0} \end{aligned}$

But still wondering: what is the primitive of $\frac{t^2}{t^2+4}$ ?

Peter van der Linden - 4 years, 2 months ago

$\begin{aligned} \displaystyle \int \dfrac{t^2}{t^2+4} \,dt &= \int \dfrac{{\left( \frac t2 \right)}^2}{1+{\left( \frac t2 \right)}^2} \,dt \\ &= 2 \int \dfrac{u^2}{1+u^2} \,du & \small \text{where } u = \dfrac t2 \implies \dfrac{dt}{2} = du \\ &= 2 \int \tan^2 \theta \,d\theta & \small \text{where } \theta = \arctan (u) \implies u = \tan \theta \implies du = \sec^2 \theta \,d\theta \\ &= 2 \int (\sec^2 \theta - 1 ) \,d\theta \\ &= 2 ( \tan \theta - \theta ) + C \\ &= 2 ( u - \arctan (u) ) + C \\ &= 2 \left[ \dfrac t2 - \arctan \left( \dfrac t2 \right) \right] + C \\ &= \boxed{t - 2 \arctan \left( \dfrac t2 \right) + C} \end{aligned}$

Tapas Mazumdar - 4 years, 2 months ago

Tnx. I knew it was a trigonometric function, but didn't know which 1 . by heart anymore.

Peter van der Linden - 4 years, 2 months ago

it's t-2arctan(t/2)

Anthony Holm - 4 years, 2 months ago

Fundamental theorem of integral calculus

The answer is 0.

1 solution

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