current attempt in progress\nestimate ∫₀¹.⁵ e⁻ˣ² dx using n = 5 rectangles to form a\n(a) left - hand…

Explanation:

Step1: Calculate the width of each sub - interval

The interval is $[a,b]=[0,1.5]$ and $n = 5$. The width of each sub - interval $\Delta x=\frac{b - a}{n}=\frac{1.5-0}{5}=0.3$.

Step2: Find the left - hand sum

The left - hand endpoints of the sub - intervals are $x_0 = 0,x_1=0.3,x_2 = 0.6,x_3=0.9,x_4 = 1.2$. The left - hand sum $L_5=\sum_{i = 0}^{4}f(x_i)\Delta x$, where $f(x)=e^{-x^{2}}$. $f(x_0)=e^{-0^{2}}=1$, $f(x_1)=e^{-(0.3)^{2}}\approx0.914$, $f(x_2)=e^{-(0.6)^{2}}\approx0.698$, $f(x_3)=e^{-(0.9)^{2}}\approx0.447$, $f(x_4)=e^{-(1.2)^{2}}\approx0.237$. $L_5=\Delta x(f(x_0)+f(x_1)+f(x_2)+f(x_3)+f(x_4))=0.3(1 + 0.914+0.698+0.447+0.237)=0.3\times3.396 = 1.019$.

Step3: Find the right - hand sum

The right - hand endpoints of the sub - intervals are $x_1 = 0.3,x_2=0.6,x_3 = 0.9,x_4=1.2,x_5 = 1.5$. The right - hand sum $R_5=\sum_{i = 1}^{5}f(x_i)\Delta x$. $f(x_1)=e^{-(0.3)^{2}}\approx0.914$, $f(x_2)=e^{-(0.6)^{2}}\approx0.698$, $f(x_3)=e^{-(0.9)^{2}}\approx0.447$, $f(x_4)=e^{-(1.2)^{2}}\approx0.237$, $f(x_5)=e^{-(1.5)^{2}}\approx0.105$. $R_5=\Delta x(f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5))=0.3(0.914 + 0.698+0.447+0.237+0.105)=0.3\times2.401=0.720$.

Answer:

(a) $1.019$ (b) $0.720$