Integral(integral exy dy from x to 1)dx from 0 to 1

Question

Stephanie Mann · Accepted Answer

Step 1  To compute the double integral.  Step 2  Given:  ∫01[∫x1exydx]dy  Step 3  Compute the integral as,  ∫01[∫x1exydx]dy=∫01[exy1y]x1dy  =∫01y(e1y−exy)dy  =∫01e1yy−exyydy  I1=∫01e1yydy  =[12e1yy2−∫−12e1ydy]01  =diverges  ∫01[∫x1exydx]dy=÷er≥s  Step 4  Thus, the integral is divergent.

xleb123 · Accepted Answer

Step 1: Evaluate the inner integral with respect to

y

:

\int_{x}^{1} e^{\frac{x}{y}} d y & = {[y e^{\frac{x}{y}}]}_{x}^{1} (Using the fundamental theorem of calculus) & = 1 \cdot e^{\frac{x}{1}} - x \cdot e^{\frac{x}{x}} & = e^{x} - x e .

Step 2: Integrate the result from Step 1 with respect to

x

:

\int_{0}^{1} (e^{x} - x e) d x & = {[\int e^{x} d x - \int x e d x]}_{0}^{1} & = {[e^{x} - \frac{x^{2}}{2}]}_{0}^{1} & = e^{1} - \frac{1^{2}}{2} - (e^{0} - \frac{0^{2}}{2}) & = e - \frac{1}{2} - (1 - 0) & = e - \frac{1}{2} - 1 & = e - \frac{3}{2} .

Therefore, the value of the given integral

\int_{0}^{1} (\int_{x}^{1} e^{\frac{x}{y}} d y) d x

is

e - \frac{3}{2}

.

Andre BalkonE · Accepted Answer

Result: Integral is 1.Solution:First, let&#039;s integrate with respect to y:∫x1exydyUsing the substitution u=xy, we can find du=−xy2dy.When y=x, u=xx=1, and when y=1, u=x1=x.Now, we can rewrite the integral as:−∫1xeuduIntegrating with respect to u, we have:[−eu]1xSubstituting the limits, we get:−(ex−e1)=−(ex−e)Now, we need to integrate this expression with respect to x:∫01−(ex−e)dxIntegrating term by term, we have:[−ex]01+[ex]01Substituting the limits, we get:−(e1−e0)+e=−e+e0+e=−e+1+e=1Therefore, the value of the given integral is 1.

fudzisako · Accepted Answer

To solve the integral ∫01(∫x1exydy)dx, we&#039;ll evaluate the inner integral first and then the outer integral.Let&#039;s begin by solving the inner integral:∫x1exydyTo evaluate this integral, we can use a change of variables. Let&#039;s substitute u=xy. This implies du=−xy2dy, which we can rearrange as dy=−y2xdu.Substituting these values, the integral becomes:∫x1exydy=∫x1xxeu(−y2x)du=−1x∫1xy2euduNow, let&#039;s solve the remaining integral:∫1xy2euduThis integral is straightforward to evaluate. We can integrate y2 with respect to u and then apply the limits:∫1xy2eudu=[y2eu]1x=x2ex−eSubstituting this result back into the previous expression, we have:−1x∫1xy2eudu=−1x(x2ex−e)=−xex+exNow, let&#039;s calculate the outer integral:∫01(−xex+ex)dxTo evaluate this integral, we&#039;ll integrate each term separately:∫01−xexdx+∫01exdxFor the first integral, we can use integration by parts. Let u=x and dv=exdx. Then, du=dx and v=ex. Applying the integration by parts formula:∫udv=uv−∫vduwe have:∫01−xexdx=[−xex]01−∫01(−ex)dx=−e+∫01exdx=−e+[ex]01=−e+e−1=−1For the second integral, we have:∫01exdx=e∫011xdxThis integral diverges at x=0, so we need to evaluate it as an improper integral:e∫011xdx=elima→0+∫a11xdx=elima→0+[ln|x|]a1=elima→0+(ln1−lna)=elima→0+(−lna)The limit lima→0+(−lna) is equal to infinity, so the second integral diverges.Therefore, the value of the given integral is −1.

Integral(integral e^{x/y} dy from x to 1)dx from 0 to 1

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