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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.8
Evaluate the following definite integrals as limit of sums.
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Question : 174 of 271
Marks: +1, -0
∫04(x+e2x)dx\int\limits_{0}^{4}(x+e^{2x})dx
Solution:  
Let f(x)=x+e2xf(x) = x + e^{2x} ,
⇒∫04f(x)dx=∫04(x+e2x)dx\Rightarrow \int\limits_{0}^{4} f(x) dx = \int\limits_{0}^{4}(x + e^{2x})dx
a=0,b=4a = 0, b = 4 and nh=b−a=4−0=4nh = b - a = 4 - 0 = 4
Now, ∫abf(x)=lim⁡h→0h[f(a)+f(a+h)+f(a+2h)+⋯+f(a+(n−1)h)]\int\limits_{a}^{b} f(x) = \lim\limits_{h \to 0} h[f(a) + f(a+h) + f(a+2h) + \dots + f(a+(n-1)h)]
=lim⁡h→0h[f(0)+f(h)+f(2h)+⋯+f((n−1)h)]= \lim\limits_{h \to 0} h[f(0) + f(h) + f(2h) + \dots + f((n-1)h)]
=lim⁡h→0h[(0+e0)+(h+e2h)+(2h+e4h)+⋯+((n−1)h+e2(n−1)h)]= \lim\limits_{h \to 0} h[(0+e^0) + (h+e^{2h}) + (2h+e^{4h}) + \dots + ((n-1)h + e^{2(n-1)h})]
=lim⁡h→0h[h⋅(n−1)(1+(n−1))2+e2h⋅1⋅(e2nh−1)e2h−1]= \lim\limits_{h \to 0} h\left[ h \cdot \frac{(n-1)(1+(n-1))}{2} + e^{2h} \cdot \frac{1 \cdot (e^{2nh}-1)}{e^{2h}-1} \right]
=4×42+12⋅e0log⁡e(e8−1)= \frac{4 \times 4}{2} + \frac{1}{2} \cdot \frac{e^0}{\log e}(e^8 - 1)
=e8+152= \frac{e^8+15}{2}
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