Math, asked by monjyotiboro, 4 days ago

Topic : integration

I found it 1/2e(e^4-1) ​

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Answered by amansharma264
3

EXPLANATION.

\sf \implies \displaystyle \int\limits^2_0 {e^{(2x - 1)} } \, dx

As we know that,

Using substitution method in the equation, we get.

Let, we assume that,

⇒ 2x - 1 = t.

Differentiate both sides w.r.t x, we get.

⇒ 2dx = dt.

⇒ dx = dt/2.

Put the values in the equation, we get.

\sf \implies \displaystyle \int\limits^2_0 \bigg[ \dfrac{e^{t} }{2} \bigg] dt

\sf \implies \displaystyle \dfrac{1}{2} \int\limits^2_0 {e^{(t)} } \, dt

\sf \implies \displaystyle \bigg[ \dfrac{e^{t} }{2} \bigg]_{0}^{2}

Put the value of t = 2x - 1 in the equation, we get.

\sf \implies \displaystyle \bigg[ \dfrac{e^{(2x - 1)} }{2} \bigg]_{0}^{2}

In Definite integrals,

First we put the upper limit then we put the lower limit in the equation, we get.

\sf \implies \displaystyle \bigg[ \dfrac{e^{(2(2) - 1)} }{2} \bigg] - \bigg[ \dfrac{e^{(2(0) - 1)} }{2} \bigg]

\sf \implies \displaystyle \bigg[ \dfrac{e^{3} }{2} \bigg] - \bigg[ \dfrac{e^{-1} }{2} \bigg]

\sf \implies \displaystyle \bigg[ \dfrac{e^{3}- e^{-1}  }{2} \bigg]

\sf \implies \displaystyle \int\limits^2_0 {e^{(2x - 1)} } \, dx = \bigg[ \dfrac{e^{3}- e^{-1}  }{2} \bigg]

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