Question # Evaluate the line integral \int_c xyds Where c is the given curve c:x=t^2,y=2t,0\leq t\leq5

Integrals
ANSWERED Evaluate the line integral
$$\int_c xyds$$
Where c is the given curve
$$c:x=t^2,y=2t,0\leq t\leq5$$ 2021-06-08
Explanation
Line integrals can be used to eavaluate integrals of two or three dimensioal curves
The curve c is expressed by the following parametric equations given by
$$x=t^2$$
and
$$y=2t$$
Thus, to evaluate the following integral given by
$$I=\int_c xyds$$
we use the following line integral formula given by
$$\int_c f(x,y)ds=\int_a^b f(x(t),y(t))|r'|dt$$
$$=\int_a^b f(x(t),y(t))|\frac{dr}{dt}|dt$$
$$=\int_a^b f(x(t),y(t))\sqrt{(\frac{dx}{dt})2+(\frac{dy}{dt})^2}dt$$
Since $$x(t)=t^2$$ and $$y(t)=2t$$
We want to evaluate $$\int_c xyds$$ as follows
Firstly, we must find the integrand $$f(x(t),y(t))$$ as follows
$$f(x(t),y(t))=xy$$
$$=t^2(2t)$$
$$=2t^3$$
Also, the derivatives of x(f) and y(t) must be
$$\frac{d}{dt}x(t)=\frac{d}{dt}(t^2)=2t$$
and
$$\frac{d}{dt}y(t)=\frac{d}{dt}(2t)=2$$
Since $$0\leq t\leq5$$, so that the upper and lower limits of the integral must be
$$a=0$$
and
$$b=5$$
Thus, using the line integral formula, we get
$$\int_c xyds=\int_a^b f(x(t))\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$$
$$=\int_0^5(2t^3)\sqrt{(2t)^2+(2)^2}dt$$
$$=2\int_0^5t^3\sqrt{4t^2+4}dt=2\int_0^5 t^3\sqrt{4(t^2+1)}dt$$
$$=2\int_0^5 2t^3\sqrt{(t^2+1)}dt$$
$$=4\int_0^5 t^3\sqrt{(t^2+1)}dt$$
To evaluate the following integral $$\int_0^5 t^3\sqrt{(t^2+1)}dt$$, we use the method of substitution as follows
Let $$u=t^2+1$$, so that
Differentiate both sides of the following equation $$u=t^2+1$$ implies that
$$du=2tdt$$
and thus
$$tdt=\frac{du}{2}$$
Also, we have
$$t^2=u-1$$
Also, we find that the upper and lower limits of the integral must be
$$t=0\Rightarrow u=0^2+1=1$$
and
$$t=5\Rightarrow u=5^2+1=26$$
Thus, the line integral becomes
$$\int_c xyds=4\int_{t=0}^5 t^2\sqrt{t^2+1}tdt$$
$$=4\int_{u=1}^{26}(u-1)\sqrt{u}(\frac{du}{2})$$
$$=\frac{4}{2}\int_{u=1}^{26}(u-1)\sqrt{u}du$$
$$=2\int_{u=1}^{26}(u^{\frac{3}{2}}-u^\frac{1}{2})du$$
Thus, evaluating the following line integral implies that
$$\int_c xyds=2[\frac{u^{\frac{5}{2}}}{\frac{5}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}]_{u=1}^{26}$$
$$=2[\frac{2u^{\frac{5}{2}}}{5}-\frac{2u^{\frac{3}{2}}}{3}]$$
$$=2[(\frac{2(26)^\frac{5}{2}}{5}-\frac{2(26)^\frac{3}{2}}{3})-(\frac{2(1)^\frac{5}{2}}{5}-\frac{2(1)^\frac{3}{2}}{3})]$$
$$=\frac{8}{15}(949\sqrt{26}+1)$$
Result:
We have shown that the following line integral implies that
$$\int_c xyds=\frac{8}{15}(949\sqrt{26}+1)$$
for the curve c expressed by the following parametric equations given by
$$x=t^2$$
and $$y=2t$$