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प्रश्न
Evaluate the following definite integrals: `int_0^1 (x^2 + 3x + 2)/sqrt(x)dx`
उत्तर
Let, I = `int_0^1 (x^2 + 3x + 2)/sqrtxdx`
= `int_0^1[x^2/sqrtx + (3x)/sqrtx + 2/sqrtx]dx`
= `int_0^1[x^2/x^{1/2} + (3x)/x^{1/2} + 2/x^{1/2}]dx`
= `int_0^1[x^{3/2} + 3x^{1/2} + 2/sqrtx]dx`
I = `int_0^1 x^{3/2}dx + 3int_0^1 x^{1/2}dx + 2int_0^1 1/sqrtxdx`
= `[x^{5/2}/(5/2)]_0^1 + 3[x^{3/2}/(3/2)]_0^1 + 2[2sqrtx]_0^1`
= `[1^{5/2}/(5/2) - 0^{5/2}/(5/2)] + 3[1^{3/2}/(3/2) - 0^{3/2}/(3/2)] + 2[2sqrt1 - 2sqrt0]`
I = `[1 xx 2/5] + 3[1 xx 2/3] + 2[2 xx 1 - 2 xx 0]`
I = `2/5 + 3 xx 2/3 + 2 xx 2`
= `2/5 + 6/3 + 4 = (6 + 30)/15 + 4 = (6 + 30 + 60)/15 = 96/15 = 32/5`
∴ I = `32/5`
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