40:T10fd,

\int_0^{\pi} \sin x \, dx = \left[-\cos x\right]_0^{\pi} = -\cos\pi + \cos 0

3d:["$","li","1",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":2}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$40"}}]]}] 41:T561,

= -(-1) + 1 = 2

— option (C).3e:["$","li","2",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":3}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$41"}}]]}] 42:T8e3,Option (A) is what you get from

\int_0^{2\pi}

, where the negative half-cycle cancels the positive one. Over

[0,\pi]

the curve stays above the axis, so the answer must be positive.3f:["$","li","3",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":4}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$42"}}]]}]

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