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Figure 1
Figure 1 shows part of the curve with equation 
y=e^((1)/(3^(2))) for 
x >= 0
The finite region 
R, shown shaded in Figure 1, is bounded by the curve, the 
y-axis, the 
x-axis, and the line with equation 
x=2
The table below shows corresponding values of 
x and 
y for 
y=e^((1)/(x^(2)))





x
0
0.5
1
1.5
2



y
1

e^(0.05)

e^(0.2)

e^(0.45)

e^(0.8)




(a) Use the trapezium rule, with all the values of 
y in the table, to find an estimate for the area of 
R, giving your answer to 2 decimal places.
(3)
(b) Use your answer to part (a) to deduce an estimate for
(i) 
int_(0)^(2)(4+e^((1)/(x^(2))))dx
(ii) 
int_(1)^(3)e^((1)/(5)(x-1)^(2))dx
giving your answers to 2 decimal places.
a)

Figure 11 Figure 11 shows part of the curve with equation y=e132 y = e^{\frac{1}{3^2}} for x0 x \geq 0 The finite region R R , shown shaded in Figure 11, is bounded by the curve, the y y -axis, the x x -axis, and the line with equation x=2 x = 2 The table below shows corresponding values of x x and y y for y=e1x2 y = e^{\frac{1}{x^2}} \newlinex x \newline00\newline00.55\newline11\newline11.55\newline22\newliney y \newline11\newlinex0 x \geq 0 11\newlinex0 x \geq 0 22\newlinex0 x \geq 0 33\newlinex0 x \geq 0 44\newline(a) Use the trapezium rule, with all the values of y y in the table, to find an estimate for the area of R R , giving your answer to 22 decimal places. (33) (b) Use your answer to part (a) to deduce an estimate for (i) x0 x \geq 0 77 (ii) x0 x \geq 0 88 giving your answers to 22 decimal places. a)

Full solution

Q. Figure 11 Figure 11 shows part of the curve with equation y=e132 y = e^{\frac{1}{3^2}} for x0 x \geq 0 The finite region R R , shown shaded in Figure 11, is bounded by the curve, the y y -axis, the x x -axis, and the line with equation x=2 x = 2 The table below shows corresponding values of x x and y y for y=e1x2 y = e^{\frac{1}{x^2}} \newlinex x \newline00\newline00.55\newline11\newline11.55\newline22\newliney y \newline11\newlinex0 x \geq 0 11\newlinex0 x \geq 0 22\newlinex0 x \geq 0 33\newlinex0 x \geq 0 44\newline(a) Use the trapezium rule, with all the values of y y in the table, to find an estimate for the area of R R , giving your answer to 22 decimal places. (33) (b) Use your answer to part (a) to deduce an estimate for (i) x0 x \geq 0 77 (ii) x0 x \geq 0 88 giving your answers to 22 decimal places. a)
  1. Identify values: Identify the values of y y from the table.\newlineValues: y0=1 y_0 = 1 , y1=e0.05 y_1 = e^{0.05} , y2=e0.2 y_2 = e^{0.2} , y3=e0.45 y_3 = e^{0.45} , y4=e0.8 y_4 = e^{0.8} .
  2. Calculate width: Calculate the width of each interval, h h .\newlineh=204=0.5 h = \frac{2 - 0}{4} = 0.5 .
  3. Apply trapezium rule: Apply the trapezium rule formula:\newlineAreah2(y0+2(y1+y2+y3)+y4) \text{Area} \approx \frac{h}{2} \left( y_0 + 2(y_1 + y_2 + y_3) + y_4 \right)
  4. Substitute values: Substitute the values into the formula:\newlineArea0.52(1+2(e0.05+e0.2+e0.45)+e0.8) \text{Area} \approx \frac{0.5}{2} \left( 1 + 2(e^{0.05} + e^{0.2} + e^{0.45}) + e^{0.8} \right)
  5. Simplify expression: Simplify the expression:\newlineArea0.25(1+2(e0.05+e0.2+e0.45)+e0.8) \text{Area} \approx 0.25 \left( 1 + 2(e^{0.05} + e^{0.2} + e^{0.45}) + e^{0.8} \right)
  6. Calculate values: Calculate the values of e0.05 e^{0.05} , e0.2 e^{0.2} , e0.45 e^{0.45} , and e0.8 e^{0.8} :\newlinee0.051.0513 e^{0.05} \approx 1.0513 \newlinee0.21.2214 e^{0.2} \approx 1.2214 \newlinee0.451.5683 e^{0.45} \approx 1.5683 \newlinee0.82.2255 e^{0.8} \approx 2.2255
  7. Substitute values back: Substitute these values back into the formula:\newlineArea0.25(1+2(1.0513+1.2214+1.5683)+2.2255) \text{Area} \approx 0.25 \left( 1 + 2(1.0513 + 1.2214 + 1.5683) + 2.2255 \right)
  8. Simplify expression inside: Simplify the expression inside the parentheses:\newline1+2(1.0513+1.2214+1.5683)+2.22551+2(3.841)+2.22551+7.682+2.225510.9075 1 + 2(1.0513 + 1.2214 + 1.5683) + 2.2255 \approx 1 + 2(3.841) + 2.2255 \approx 1 + 7.682 + 2.2255 \approx 10.9075
  9. Multiply by 00.2525: Multiply by 00.2525:\newlineArea0.25×10.90752.73 \text{Area} \approx 0.25 \times 10.9075 \approx 2.73
  10. Answer for part (a): Answer for part (a):\newlineArea2.73 \text{Area} \approx 2.73
  11. Deduce estimate for integral: Use the result from part (a) to deduce the estimate for 02(4+e1x2)dx \int_{0}^{2} (4 + e^{\frac{1}{x^2}}) \, dx :\newline02(4+e1x2)dx4×2+2.73=8+2.73=10.73 \int_{0}^{2} (4 + e^{\frac{1}{x^2}}) \, dx \approx 4 \times 2 + 2.73 = 8 + 2.73 = 10.73
  12. Deduce estimate for integral: Use the result from part (a) to deduce the estimate for 13e15(x1)2dx \int_{1}^{3} e^{\frac{1}{5(x-1)^2}} \, dx :\newlineSince the integral is over a different interval and function, we cannot directly use the result from part (a). This part requires a different approach.

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