# State True or false. (a) integral_{-1}^1 (x^5 - 6 x^9 + {sin x} / {(1 + x^4)^2}) dx = 0. (b) If...

## Question:

State True or false.

(a) {eq}\displaystyle \int_{-1}^1 \bigg(x^5 - 6 x^9 + \dfrac {\sin x} {(1 + x^4)^2}\bigg) dx = 0 {/eq}.

(b) If {eq}\displaystyle f(x) = \int_0^x \sin (t^2)\ dt, \ f'(x) = x \sin (x^2) {/eq}.

(c) If {eq}f' {/eq} is continuous on {eq}\displaystyle [a,\ b],\ 2 \int_a^b f(x)\ f'(x)\ dx = [f(b)]^2 - |f(a)]^2 {/eq}.

(d) {eq}\displaystyle \int f(x)\ dx = x\ f(x) - \int x f'(x)\ dx {/eq}.

## Differentiation of Integral with limits:

The finite integral can be differentiated as follows,

{eq}\dfrac {d}{dx} \int_0^{h(x)} f(t) dt=\dfrac {d}{dx} \left[f(h(x))\right]\\ \dfrac {d}{dx} \int_0^{h(x)} f(t) dt=h'(x) f(h(x)) {/eq}

For example, {eq}\dfrac {d}{dx} \int_0^{x^2} sin(t) dt=\dfrac {d}{dx} \left[sin(x^2)\right]=2x \ \sin(x^2) {/eq}

(a) {eq}\displaystyle \int_{-1}^1 \bigg(x^5 - 6 x^9 + \dfrac {\sin x} {(1 + x^4)^2}\bigg) dx = 0 {/eq}.

That is true since this function {eq}\bigg(x^5 - 6 x^9 + \dfrac {\sin x} {(1 + x^4)^2}\bigg) {/eq} is an even function integrated over {eq}[-1,1].{/eq}

(b) If {eq}\displaystyle f(x) = \int_0^x \sin (t^2)\ dt, \ f'(x) = x \sin (x^2) {/eq}.

That is false. Since, {eq}\dfrac {d}{dx} \int_0^x \sin (t^2)\ dt, \ f'(x) = \sin (x^2){/eq}

(c) If {eq}f' {/eq} is continuous on {eq}\displaystyle [a,\ b],\ 2 \int_a^b f(x)\ f'(x)\ dx = [f(b)]^2 - |f(a)]^2 {/eq}.

That is true.

{eq}2 \int_a^b f(x)\ f'(x)\ dx = 2 \int_a^b f(x)\ \dfrac {df(x)}{dx} \ dx \\ = 2 \int_a^b \ \dfrac {1}{2} \dfrac {df(x)^2}{dx} \ dx \\ = \int_a^b \ df(x)^2 \\ =\left[f(x)^2\right]_a^b\\ =[f(b)]^2 - |f(a)]^2\\ {/eq}.

(d) {eq}\displaystyle \int f(x)\ dx = x\ f(x) - \int x f'(x)\ dx {/eq}.

That is true. Since, {eq}d(uv)=udv +v du\\ u dv=d(uv)-v du\\ \int u dv=uv- \int v du\\ {/eq}

For {eq}\displaystyle \int f(x)\ dx {/eq}

let {eq}dv=dx, \ u=f(x){/eq}

Hence, {eq}v=x\\ \displaystyle \int f(x)\ dx= x\ f(x) - \int x f'(x)\ dx {/eq}.