The correct option is B 2 units I=∫^π/2_ π/2cos x dx=sin x |_ π/2^π/2=sinπ/2 sin( π/2)=1+1=2 ∵∫_a^b f(x)dx=I(x) |_a^b=I(b) I(a) ...

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Definite Integrals

∫-π / 2π / 2c...

Question

$\int_{- π / 2}^{π / 2} c o s x d x$

1 unit

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2 units

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0 unit

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None of these

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Solution

The correct option is B
2 units

$I = \int_{- π / 2}^{π / 2} c o s x d x = s i n x |_{- π / 2}^{π / 2} = s i n \frac{π}{2} - s i n (- \frac{π}{2}) = 1 + 1 = 2 ∵ \int_{a}^{b} f (x) d x = I (x) {∣ ∣ ∣}_{a}^{b} = I (b) - I (a)$

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\frac{π}{2} I

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